ok, let’s see, the output transformer is there to match the high impedance/low current output of the power tube to the low impedance/high current load of the speaker. Note that the larger leads are on the secondary side of the transformer where the higher current is. In matching the power tube to the speaker load, you need to know what that primary impedance of the transformer is OR what the speaker load is.
You know the speaker load is 8 ohms and you know a single 6V6 (or 6L6)
in a class A circuit (like the Two Stroke) likes around 5000 ohms, so to find the “turns ratio” you use Ohm’s law: I=E/R. As an example if you apply a AC voltage to the primary side of the transformer, say 5v AC. Now measure the AC voltage of the secondary, say we get .21v AC. The ratio is 5 / .21 = 23.8, so that a turns radio of 24:1. The impedance ratio is the square of the turns ratio so: 24 x 24 = 576, Now multiply the impedance ratio (576)by the load (speaker) impedance (8 ohms): 8×576 and you get 4608. That’s about 4.5K which is close enough to 5K to work with a 6V6.
The bottom line is, the 4,8 and 16ohm taps of the output transformer are only accurate if you know the primary impedance of the OT. The transformer supplier should be able to provide it. Once you confirm that the primary for your OT is 4500-5500, you should be good to go with the 8 ohm tap. Clear?