Maxwell's Demon: Why does the entropy of the overall system decrease?
You can think of entropy as measuring the amount of phase space volume that the system could be occupying. (This is the set of all possible sets of positions and momenta for all the particles.)
It's correct that the set of possible momenta are unchanged, so that doesn't affect the entropy. But the set of possible positions is lower, because it becomes correlated with momentum -- the high-momentum particles can only be in half the volume as before. That's why the entropy decreases.
You also specifically asked how the entropy of side $B$ changes. This depends on how the system is set up. If you neglect particle interactions, then side $B$ ends up with only fast particles, while it began with both fast and slow particles. Since the available phase space is lower, the entropy of side $B$ decreases.
However, if you allow particle interactions, the particles in side $B$ will reach thermal equilibrium, and you'll get a distribution of velocities (i.e. some slow, some fast, and some very fast). Then the entropy of side $B$ increases.
The first way is more common in popular presentations since it's a bit cleaner, while the second is more realistic. But in both cases, the total entropy of $A$ and $B$ decreases, which is the key point.
So the box is hot on the right and cold on the left. Then one can use this separation of temperature to run a heat engine by allowing the heat to flow from the hot side to the cold side. Another possible action of the demon is that he can observe the molecules and only open the door if a molecule is approaching the trap door from the right. This would result in all the molecules ending up on the left side. Again this setup can be used to run an engine. This time one could place a piston in the partition and allow the gas to flow into the piston chamber thereby pushing a rod and producing useful mechanical work.
So remembering this is a paradox, if the demon "could" do his job, a system that previously could do no work because entropy was even all over, is now able to do work, so by definition entropy has decreased.
But the demon can't do the job, so this is just assuming he could.
Using the expression for the internal energy of an ideal gas, the entropy may be written:
$${{\frac {S}{Nk}}=\ln \left[{\frac {V}{N}}\,\left({\frac {U}{{\hat {c}}_{V}kN}}\right)^{{\hat {c}}_{V}}\,{\frac {1}{\Phi }}\right]}$$
Since this is an expression for entropy in terms of $U$, $V$, and $N$, it is a fundamental equation from which all other properties of the ideal gas may be derived. Note, no explicit $T$ dependence exists, as mentioned in the comments above.
There are different kinds of entropy here. The thermal entropy due to the molecules moving faster or slower is a red herring and actually isn't important for understanding Maxwell's demon at all. The important entropy here is the mixing entropy that has to do with how well-sorted these different kinds of molecules are.
Personally, I think this version of Maxwell's demon shouldn't be taught in physics classes at all, even though it's the way in which Maxwell originally formulated it. A much more straightforward way of explaining Maxwell's demon is to just talk about different kinds of molecules like Nitrogen vs Oxygen. (An even simpler version is to just call them blue and red balls, although this may lead people to ask how balls tiny enough to be considered "microscopic" can have distinct colors.)
Whether blue and red, Nitrogen and Oxygen, or fast and slow, the important point is just that there are 2 different types of particles. And when they are all mixed together this is a state of high entropy, whereas when they are carefully separated into different chambers this is a state of low entropy.