DC analysis of common source MOSFET circuits to find gate voltage using different power supplies
The Vg eqn for the 1st circuit is correct.

The 2nd circuit has a negative rail (also called Vdd) and this is where the confusion lies. You could regard the voltage at the gate as being twice the 1st circuit but relative to -5V.

Vdd-(-Vdd) = 2*Vdd = 10V because Vdd on its own is 5V. This and the standard potential divider formula gives you a gate voltage relative to the -5V rail.

But, because the gate voltage is more sensibly referenced to 0V, then Vdd needs to be subtracted to make Vg relative to 0V rather than -5V.