The resistor used should be even be higher but is impractical due to the max output swing one would get.
The resistors need to be placed in series.
A current source has an infinite (extremely high) output resistance which will be emulated by a high series resistance.
These resistors will just give you a glimpse what to expect, real current drive is even more exaggerating.

As said before (by quite a few) HP and speakers are designed to be driven from a voltage.
On the other side a lot of people like raised bass (as long as it doesn't become fat/muddy) and may prefer current drive simply as it sounds better to them (moar bass).
It is 'fake' though and could just as well turn up a bass slider of a multiband EQ with a normal (low distortion) amplifier and raise the treble a dB or 2.

I got all of that... But one of the things that is sort of weird from the Bakoon clusteramp is that I was able to get 9.78 Vrms in current mode into 300 ohms. In voltage mode I was only able to get 3.39 Vrms into 300 ohms.

The output voltage has a limit as well (just like any other amp) which is about 10V in this case.
The power supply rails voltage thus will probably be +/-15V and the current limit is about 140mA (peak = 100mA_RMS)

It could be that in voltage mode the circuit is changed so much by the different feedback topology that this limits the output voltage in that mode.
It may also be (due to the amps topology) that to reach an output voltage above the 3V perhaps you need a higher input current/voltage (higher than the 2i2 provides)
Maybe drive it from the headphone out (if that has a higher voltage swing) or an amplifier that can provide a higher voltage.
Most likely both different outputs have completely different circuits with different limits.

Follow the series resistance + series connection stuff for voltage source type amps.

In the Bakoon maybe they are doing some resistor thing to emulate voltage out instead and maybe that's why things don't look so hot from that port...??

EDIT: OK, actually I don't think one can get the impedance effect with a small resistor in parallel to the driver when using voltage drive... but when using current drive one might... Still dunno exactly what the Bakoon amp is all about (how it works and all I mean).

They simply change the feedback point to achieve voltage gain.

When you parallel the output of the current section with a small value resistor it effectively lowers the impedance peak and thus the FR will become flatter BUT the output voltage will also be lower as part of the output current will flow in the resistor.
So yes, paralleling a resistor will make a current amp act more like a voltage amp (when small enough in a relative sense) while mounting a resistor in series with a voltage amp will make it act as a current amp.

To make a current amp act exactly like a voltage amp the parallel resistor should be < 1 Ohm and thus it will have almost no output voltage.
To make a voltage amp act exactly like a current amp the series resistor should be >100kOhm and thus it will provide almost no output current.

It appears to be a current amplifier without feedback, so not exactly the same as shown in the drawings I posted.
It looks like the 'heart' of the circuit is very similar to the inverting input of an 'opamp' (with the + input connected to ground) and the voltage follower output stage has been replaced by a current output stage. All internal stages are likely to have local feedback instead of overall feedback acc. to their vague description.
The circuit simply amplifies the input current. As transistors amplify currents in a rather linear way it is easy to make such a circuit perform nicely.
In a voltage input amp the input is voltage controled (as well as the output is a voltage control) and this has to be achieved with current amplifying devices.
FET's and tubes are input voltage and output current devices.

In voltage mode they may simply take their feedback point from another part inside the circuit.

do BOTH outputs work simultaniously or as soon as you plug one headphone in the other output is disabled ?

The current mode out also puts the whole 'damping factor/grip' thing and output resistance debate in a different light though.
Given the fact that a current out amp doesn't have a damping factor at all and output resistance is in the M Ohm range.

Cable resistance and inductance also do not play a role any more (in current mode) and cable capacitance will only become relevant far above the audible range.

I like to think of it as a novelty amp... it's not like everyone is suddenly going to make current out speaker- and headphone-amps all of a sudden as this would be 'new insights'.
The current source concept is already very old and not applied because headphones/speakers are designed to be voltage driven.
Even the old 5 pole DIN connector used 'current out' for the recording pins and voltage out for the output pins.
More of a gimmick than a real attempt to improve fidelity (although the manufacturer may be convinced it is).

I'll stick to voltage out amps myself with some 'current output alike' output customer selectable settings as well though.

Quote (selected)

To make a voltage amp act exactly like a current amp the series resistor should be >100kOhm and thus it will provide almost no output current.

Solderdude: I don't understand how this would make a voltage amp supply any current through the load. How will it behave in a current drive mode with a huge resistance choking the current prior to the speaker/phones?

It's only theoretical hence the words: thus it will provide almost no output current.

IF you were to convert a voltage amp to a current amp the voltage rails would have to be very high in order to provide a current.
With 100kOhm the voltage rails would have to be a few kV in order to obtain practical values for headphones so not realsitic in any sense.

Of course using a 1k Ohm resistor in series with the output of a voltage amp will already transform it towards being a current amp.
A real current amp will have an Rout >1Mohm though (within the output voltage range it can operate in, outside of it the impedance drops dramatically).
In practice, with nominal impedances of around 30 Ohm you won't measure much differences between a real current amp and a voltage amp with 1k in series (except for output power).
For 300 Ohm headphones there will still be a difference unless you 'up' that 1k to say 10k or so.

The 100k was to make a point that converting a voltage amps into a 'real' current amp isn't really feasable in practice.

Correct me if I'm wrong, been a while since I studied Circuit analysis, if you parallel the load with a high value resistor, current will be choked through the resistor, most of the voltage will drop across the resistor as well. Thus the load will get less voltage drop, but still get the balance of the current. Output of course would be lower

If you parallel a load (assume 30 Ohm) with a high value resistor (say 300 Ohm) most of the current will flow though the 30 Ohm, after all they have the same voltage on them.
The amp will see a load of 27 Ohm and probably won't behave much differently.

Perhaps you meant:  if you put a high value resistor in series with the load most of the voltage will drop across that resistor.
That would be correct and indeed the load would get almost no voltage across it BUT the current would always be virtually the same and thus act as a current source.

Example HD558: R load = 60 Ohm (call this A) and peak R load is 300 Ohm (call this B).
With a 120 Ohm 'source' and the voltage source being 10V 
A would draw 10/180= 55.5 mA where B would be 10/420= 23.8mA so the current is not constant as Rsource is too small compared to R load differences.

Same example except R source = 10k
A would draw 10/10060= 0.99 mA where B would be 10/10300 = 0.97mA which is more 'constant'.

Same example except R source = 1M
 A would draw 10/1000060= 9.999 uA where B would be 10/1000300 = 9.997 uA which is more even more 'constant'

The higher the series R will be the small er the difference and the closer it gets to a real current source but also a lower current.
If you would like at least 10mA with a 1M R you will need 10kV input voltage.
This would be very dangerous when touching the terminals on open circuit.

The reason why the Bakoon can deliver more current while the max voltage is 10V is because of the circuit used.
The Bakoon simply isn't a voltage amp with a high output R but an actual current source.

For this reason it is likely the Bakoon has either 2 separate amp circuits (one current, one voltage) which would explain some of the findings such as THD or the current circuit has applies voltage feedback in some way or adds a fixed resistor + voltage follower for the output voltage stage.

I'm surprised it does not benefit orthos given ortho's relatively flat impedance.

That's exactly why there is no benefit (viewed objectively, so probably not while subjective determined by some).
A constant current (over the entire freq range) over a constant R (over the entire freq range) will yield a constant voltage (over the entire freq range) .
A 'normal' voltage amp will always give a constant voltage over an entire frequency range (regardless of impedance).
So both types of amps will both yield a flat frequency range with orthos.

Headphones with a varying impedance will show a flat (electrical measured) frequency response on a 'normal' voltage amp.
Headphones with a varying impedance will show a 'curved' (electrical measured) FR on a current output amp (an exact copy of the impedance plot) and thus altering the sound where with orthos (or other linear dynamics) the FR will be flat, as flat as from a voltage out amp.