M2 is essentially acting like a pullup resistor in this case. Real resistors are difficult to make on silicon chips, so a PFET in on-state is good enough for this purpose.
The chip designer can vary parameters like the channel length, width, and possibly doping level. Depending on the characteristics of the transistor, it could act more like a current source than a resistor at the operating point. Sometimes a "long tail FET" is used to make a rough current source. Without knowing the parameters of M2, we don't know if it is more like a resistor or more like a current source, although in this application that wouldn't make much of a difference. Ideally you'd want a current source for a pullup, but lots and lots of places you see resistors doing that job well enough.
You would not be able to control both series source-drain voltages simultaneously. Try to draw out this circuit, with the controlling voltage sources in place. You would need to control both FET's to have the same Vgs and Vds. Once you attempt to draw it out you will realize that it cannot happen (falstad circuit simulator is nice for this, you can probe V and I, in real time).
If you attempt to do this (set both transistors Vgs and Vds simultaneously), you will not have a series circuit, since the voltage supplies would be in parallel. In this situation, the power supplies would source/sink whatever currents were needed to make the FET's Vds equal, but equal currents would not flow through each FET.
In series, there will be more voltage drop on whichever FET is weaker. This provides equal current to the FET which is stronger (with lesser Vsd drop). If one (or both) FET must go into the linear region for this to occur, it will.
Two transistors in series with different W/L ratios combine exactly like parallel resistors. This makes sense because the W/L ratios can be considered as a conductance.
As two numerical examples, consider two similar FET's with the same W/L = 2
If these FET's are placed in series, the 'equivalent' single FET W/L is ~2//2 = 1.
Now, consider two similar FET's with W/L of 1 and 2. The equivalent W/L, if these devices are placed in series and driven with the same gate signal, is ~1//2 = 0.66
So, adding a similarly driven strong FET in series with a weaker FET just makes the chain weaker overall.
Of course there is the body effect (if these are built on-die, which is implied since you can vary the W/L), which increases the threshold voltage of whichever FET's do not have their source at the most negative/positive voltage, for NMOS and PMOS respectively. This can be thought of as reducing the W/L ratio. This occurs if you have two or more of either type in series (2+ NMOS or 2+ PMOS). A CMOS inverter does not suffer the body effect since both NMOS and PMOS have their sources at the respective supplies.
Best Answer
The width differences arise from the difference in mobility of the electrons and holes with holes being a factor of ~2.2 slower than electrons.
\$ \frac{W}{L} \mu_e = \frac{W}{L} \mu_h \$
Having balanced \$ g_m\$ on teh PMOS and NMOS will give symmetrical rising and falling edges times. Which is not a necessary condition for low power as you state. Low power design arises through other mechanisms.