From Lang's Algebra:
Let $A$ be commutative. Then $M$ is a module over $\text{End}_A(M)$. If $R$ is a subring of $\text{End}_A(M)$ then $M$ is a fortiori a module over $R$. More generally, let $R$ be a ring and let $\rho: R \rightarrow \text{End}_A(M)$ be a ring homomorphism. Then $\rho$ is called a representation of $R$ on $M$. This occurs especially if $A = K$ is a field.
Here, $A$ is a ring, $M$ is an $A$-module, and $\text{End}_A(M)$ is the ring of endomorphisms of $M$.
Why do we require $A$ to be commutative? The results still hold even if $A$ is noncommutative, no? I have let the action of $\text{End}_A(M)$ on $M$, $f \cdot m$ for all $f \in \text{End}_A(M)$ (which are $A$-homomorphisms) and $m \in M$, be defined by $f \cdot m = f(m)$. Clearly this gives a module structure for $M$ over $\text{End}_A(M)$ by the usual laws of addition and composition of homomorphisms, regardless of the commutativity of $A$. What is the issue here?
Personally, I tend to think that Lang requires $A$ to be commutative so that the operation $A\times M\rightarrow M$ given by $a\times m\rightarrow a\cdot m$ is an $A$-linear ($a$ is a fixed element in $A$). This need not to be an $A$-linear when $A$ is not commutative.