In this question, the author finds the probability mass function and then finds the expectation because it happens to be a hypergeometric and therefore the answer is well known.
In general, how do you find the expectation by first finding the probability mass function and/or moment functions?
If I wanted to use conditional expectation, would this be correct : $$E(X)=\sum_{k=0}^mE[(X|X=m-k)]P(X=m-k) = \sum E(X)P(X=m-k) $$ I am not sure if I have to sum over all y, or how this should be set up in general.
Also, if I wanted to solve this with moment generating functions, am I right that it would become $$M_{x+y}=M_xM_y=(pe^t+1-p)^n(pe^t+1-p)^n=(pe^t+1-p)^{2n}$$ Then, to get the expectation of x, evaluate $M_{x+y}'(0)$, but how would I condition the moment function? $(pe^t+1-p)^{2n}|x+y=m$ ?
Shortcut:
From this it follows directly that: $$\mathbb E[X\mid X+Y=m]=\frac{m}2$$
The first equality in your question is correct (and quite useless), but the second is not correct.
This because $\mathbb E(X\mid X=m-k)=m-k$ (not $\mathbb EX$).