Transfer function of this circuit
If Vx is the voltage at the "upper" secondary we'll have -Vx at the "lower" secondary. (You called this the input in your question but the input is in the primary).

If Vx>5.5 then both diodes are ON, one forward and the other in Zener region. The "upper" diode wants Vx at the output, the "lower" wants -Vx+11. Who wins? It depends on the resistance of each path. Assuming diodes ideals and for the transformer equal resistance in upper and lower secondary coils, we have a "tie". Then, assuming this coil resistance much smaller than R_load, the output voltage is simply the average Vo=Vx/2+(-Vx+11)/2=5.5 V.