Why do we assume VGS0 for a mosfet in cuttoff and Vce>0 for a BJT in cutoff?
A simple model for you is to think of the term cutoff to mean that the device is "cut off" or not operating or not controlling that pin (the collector in the NPN or the Drain in the NMOS).  In both of those cases it just means that the pin can take any value that it wants or is imposed upon it by other circuitry.  So it really means  that \$ V_{DS}&gt;= 0\$ and \$V_{CE} &gt;= 0 \$ i.e. any value it wants.