Electronic – Why do we assume VGS0 for a mosfet in cuttoff and Vce>0 for a BJT in cutoff

The Vth of a NMOS is the Vgs at which it operate until (ie. so long as Vgs > Vth it will continue to conduct).

As you have stated this Vth is 1.0V. Initially when the switch is closed the NMOSes will turn on as the voltage at their respective sources will be zero and the voltage at the gates will be 2.5 - this is well above the Vth that you mentioned of 1.0V. As the capacitor charges through these NMOSes which are now more or less making short circuit the voltage at the source is increasing and it will continue to do this until it reaches 1.5V at which point the Vgs will start to go below 1.0V thus making the NMOS transistors turn off.

The fact that there are two of them doesn't really make any difference, there could be 1 or 10 and it would have the same result.

The mistake you made by thinking it will only charge to 0.5V is that the voltage drop is not from the drain - source so there would not be two drops, it si only to do with the potential between the gate and source and as explained once this goes below 1.0V the transistor turns off.

\$V_{gs}\$ defines the thickness of the channel under the Gate (in a 3D mosfet model). Think of this channel as a hallway whose width you can increase as you increase \$V_{gs}\$. Think of people walking through the hallway as electrons ( \$I_{d}\$). Think of yourself as a god who can force people to have to run through this hallway (force = applied \$V_{ds}\$, which creates \$I_{d}\$)

If the hallway is wider (\$V_{gs}\$ is higher), you can fit more running people (electrons) before people start getting stuck in a tight trafficy environment (current saturation).

If you keep the hallway constant (\$V_{gs}\$ is constant), the there is only a finite amount of people (finite \$I_{d}\$) you can force (increase \$V_{ds}\$) before people start getting stuck and jammed up (\$I_{d}\$ saturates). If you try to force more people into the jammed hallway (try to increase \$I_{d}\$ by increasing \$V_{ds}\$), you will barely be able to. It will be a lot harder for you because the doorway is already jam-packed (\$I_{d}\$ = saturated) with people (current).

I like to think in analogies because all of these phenomena are already experienced by us in everyday life. This analogy gives a rough idea of whats happening under the hood. :) Hope it helps!