I have a simple conceptual question rather than an actual problem so pardon me for not using the provided template.

There's a question in my Microelectronic Circuits book involving this circuit:

where it states that

Part of the solution involves this equation:

V_{OV4} = V_{D4} - V_{S4} (equation for overdrive voltage of Q_{4})

V_{OV4} = -1.5 - (-1.3)

My question is: Why is V_{D4} equal to -1.5 V and why is V_{S4} equal to -1.3 V? -1.3 V is the voltage at the drain of Q_{5}, how did it become the voltage at the source of Q_{4}?

Unfortunately, what you state is incorrect. The drain of D4 will be more positive than the source.
The drain of Q4 is > -1.5V. Perhaps as low as -1.3V but certainly no lower.
The source of Q4 is at -1.5V.
The drain of Q5 can be as low as -1.3V.
For an NMOS. the drain is more positive than the source unless it's operated in the "inverse" mode in which case the labels "drain" and "source" become a matter of semantics. I have never encountered a circuit using the inverse mode, except once which turned out to be a design error. So forget the inverse mode.

New question related to same circuit in the initial post: the solution manual states that V_{DS2max} = V_{OV2} = V_{GS2} - V_{th}. My question is why is V_{DS2max} = V_{OV2}? Shouldn't V_{DS2} = V_{GS2} since the drain and gate of Q2 are connected?

I get that but what about the threshold voltage (V_{th})? The overvoltage is the voltage across two transistor terminals minus the threshold voltage. Why don't we include it in the equation for V_{OV} = V_{DS2max}?