How is $\text{End}(M)$ a ring? Let $G$ be an abelian group. I am told that $\text{End}(M)$ forms a ring. 
I don't see how that is. The property that I am having difficulty proving is that if $f,g\in \text{End}(M)$, then $f+g\in \text{End}(M)$. Any help would be much appreciated. 

Write $h=f+g$. What you need to show is that $h:M\to M$ is a group homomorphism, that is, you need to show that for all $m,m'$ in $M$
$$h(m+m')=h(m)+h(m')$$
Just expand both sides separately using the fact that $f$ and $g$ are by definition group homomorphisms, and rearrange the sums using the fact that $M$ is abelian, and you should get the desired equality.
Here are the details:

The first term is \begin{array}{rcl}h(m+m')&=&(f+g)(m+m')\\&=&f(m+m')+g(m+m')\\&=&f(m)+f(m')+g(m)+g(m')\end{array} and \begin{array}{rcl}h(m)+h(m')&=&(f+g)(m)+(f+g)(m')\\&=&f(m)+g(m)+f(m')+g(m')\end{array} The two expressions are equal because $(M,+)$ is abelian.

