Hi,
I see you can charge up to 4 batteries simultaneously? This will determine the output voltage.
Some thoughts, perhaps you can get some clues from the board.
On the one side it gives a voltage of 2.7V which is consitent with charging two cells. Both sides of the board seem identical (input componentry) and are fed by what appears to be identical windings. I think that indeed you have a dual circuit, one charges two AA/AAA and the other side does the same. This is to avoid sensing componentry that will automatically sense how many cells and therefore the voltage that need applying.
Ideally you need 1.4V per cell and I would suggest that the winding (red to white) is as follows;
2.8V (required for charging) + 0.7V (for the drop over the diode) + voltage drop over the 2.7ohm resistor. As this seems to be the one leg of the 9V circuit I thinks they drop the voltage through that resistor?
Can you confirm that the charger works in pairs of batteries?
Check your resistance measurements, I think that the windings are A (Red and White), B (Red and White) and we need to sort out the black wire.
The black winding I believe is to feed the 9V (8,4V) batteries. I think that this winding is also dual having the black as a common. (Red and Black and Red). The voltage of this winding should be 10.5 to 11.9V. Again I believe that each side feed one 9V battery.
By measurement I think you will get the following;
Between the Red and White on both sides, identical resistances.
Between the Black and Red on both sides identical resistances.
Can you check the resistance in both cases of black to white?
PS: I am using logic rather than any of the extremely limited electronics knowledge I may possess.
Lets take it from there.
Cheers
Andrew