The secondary may be modeled as two ideal sources for purposes of this discussion. It is done all the time.

Because separate sources have separate motive fields and are therefore not single-phase. Therefore that model is outside the scope of establishing correct phasors for a 120/240 system.

First of all, what happened to I[Rcos? + Lsin?]? And what about the arguments? You conveniently replaced that expression with I[ R + jwL] which you lifted from my response.

Not needed. Confused you. Your expression carries the important current element so I can go with that. Holding on the prior expression would drag the discussion off-topic.

And how did the current, I, find its way into this thread?

The current is required to determine the correct phasor direction.

Yes indeed, the impedance, R + jwL, carries a phase angle, as well as V and I which are also phasors.

(R+jwL) does not carry a phase angle. It's the phase deflection imposed on I passing through the secondary coil based on the direction of the current. The phasor direction for the voltage starts with the phase angle of I and is deflected to lead or lag because of (R+jwL).

The primary EMF (supply) drives the current in a unified direction (call it <0) across the secondary coil (load) from A through N to B.

V=I(2*[R+jwL])=I[sub]an[/sub][R+jwL]+I[sub]nb[/sub][R+jwL]=V[sub]an[/sub]+V[sub]nb[/sub]

Because the secondary is a load to the primary and there is only one EMF driving the current: I[sub]an[/sub]==I[sub]nb[/sub]

Because of how we construct secondary coils: [R+jwL][sub]an[/sub]==[R+jwL][sub]nb[/sub]

So now you move your reference point to N.

V[sub]nb[/sub] == -V[sub]bn[/sub] ~= V[sub]an[/sub] < 180

which is where you stop. And which is correct in field practice if all you're doing is working on voltage.

But if you wish to comment about the system phase then you have to continue:

I[sub]nb[/sub] == -I[sub]bn[/sub] ~= I[sub]an[/sub] <180

and your power resolves as:

V[sub]an[/sub]*I[sub]an[/sub]+V[sub]bn[/sub]*I[sub]bn[/sub] == V[sub]an[/sub]*I[sub]an[/sub]+V[sub]bn[/sub](<180)*I[sub]bn[/sub](<180) == 2*V[sub]an[/sub]*I[sub]an[/sub]

which means the power to the system is all single-phase.