Hi,

You are that M1 will be saturation throughout, whereas M2 is in cutoff when VG2 is 0, in saturation when VG2 is VDD/2 and in triode when VG2 is VDD.

But the output will not be exactly zero for VG2=VDD. It will be a small value close to the overdrive of M2 because some current is still flowing through this branch.
Also when both input voltages are VDD/2, ideally Vout should be VDD/2 for single ended configuration as in this picture, and Vout is zero for a fully differential output, as for a fully differential structure the output Vout = Vout1-Vout2.

So in the given picture Vout = VDD-Vov4, when VG1=VG2 = VDD/2,    where Vov4 = VGS4 - Vth4 = overdrive voltage of M4.

Attaching a simulated plot of the same. Vin and Vo are shown with currents of both branches. In the plot M0 corresponds to M1 in the picture with VG=VDD/2, and M6 from the plot is the same as M2 of picture, with VG2 being swept from 0 to VDD as indicated by the ramp.
Note the VDD here is 1V.