To the first question the answer is negative. 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. 

To the last question, the answer is yes (and it is important that $\Omega(f)=M$), by that I mean $\Omega(f)=\Omega(f^m)=M$ for every $m\geq 1$. 

The proof I know goes as follows:

Consider a basis $\{A_n\}$ of the topology of $M$ and $O_n=$ {$x\in A_n \ : \ f^j(x) \in A_n \ for \ some \ j \geq 1$ } $\cup \overline{A_n}^c$. The set $O_n$ is open (because $A_n$ is open) and dense since every point is non wandering (so, given an open set $U\subset A_n$ there exists $x\in U$ such that for $j\geq 1$ we have $f^j(x) \in U \subset A_n$). (Notice that if $\Omega(f)\neq M$ the set $O_n$ could fail to be dense in $\Omega(f)$). 

If $x\in R=\bigcap O_n$ we get that $x$ is a limit point for $f$ (in fact, it will be recurrent since given a neighborhood $U$ of $x$, there is $x\in A_n \subset U$, and since $x\notin \overline{A_n}^c$ and $x\in O_n$ we get that it has a future iterate in $A_n$)  and thus also for $f^m$. Notice that $x$ may a priori be not recurrent for $f^m$, but it will be a limit point. 

Since the [limit set][1] is contained in the nonwandering, we obtain the result. 

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


  [1]: http://en.wikipedia.org/wiki/Limit_set