To the last question, the answer is yes (and it is important that $\Omega(f)=M$), by that I mean $\Omega(f)=\Omega(f^n)=M$. The proof I know goes as follows:

Consider a basis $\{A_n\}$ of the topology of $M$ and $O_n=$ {$x\in A_n \ : \ f^n(x) \in A_n \ for \ some \ n\geq 1$ } $\cup \overline{A_n}^c$. The set $O_n$ is open (because $A_n$ is open) and dense since every point is non wandering. 

If $x\in R=\bigcap O_n$ we get that $x$ is a limit point for $f$ and thus also for $f^n$. Since the limit set is contained in the nonwandering, we obtain the result. 

This aplies also to get that if $\Omega(f|_{\Omega(f)})=\Omega(f)$ then $\Omega(f)=\Omega(f^n)$.  

To the first question the answer is negative, I believe that there are two homeomorphisms of the circle with irrational rotation number such that their composition is Morse-Smale (in fact, you can multiply two $2\times 2$ matrices with complex eigenvalues to get one hyperbolic matrix, the action on the proyective space does the trick).  This implies that $\Omega(f)=\Omega(g)=S^1$ but $\Omega(f\circ g)$ consits of two points.