Constructing an $R$-algebra from an $R$-module, and their relation Let $R$ be a commutative ring, and let $M$ be an $R$-module. Then every $r \in R$ could be seen as a group endomorphism on $M$, i.e. we have a natural map $f : R \to \operatorname{End}(M)$. This map is a ring homomorphism, and so by setting $r\cdot \varphi := f(r)\varphi$ for $\varphi \in \operatorname{End}(M)$ we have an action of $R$ on $\operatorname{End}(M)$, which gives us an $R$-algebra.
Is there anything known about the relation of this $R$-algebra to the module $M$, some theorems relating those two?
 A: 
we have an action of $R$ on $\operatorname{End}(M)$, which gives us an $R$-algebra.

No, not quite. To make $End(M)$ an $R$ algebra, the map $f$ would have to map into the center of $End(M)$.
For example, $\mathbb H$ is a $\mathbb C$ module, and has such a map, but there is no map of $\mathbb C$ into the center of $\mathbb H$. $\mathbb H$ is not a $\mathbb C$ algebra.
The map $f$ you are talking about is a well-known equivalent way of defining what it means to be an $R$ module. You can either specify all the axioms about how the addition and multiplication works, or you can say "a ring homomorphism from $R$ into $End(M_\mathbb Z)$."
As for your last question, let me say this. There is not such a tight connection between $M_R$ and $End(M_\mathbb Z)$, other than the property we mentioned that $End(M_\mathbb Z)$ contains a homomorphic image of $R$.
The really interesting relationship is between $M_R$ and $End(M_R)$. You can say lots of things about that relationship. For example, the direct summands of $M_R$ correspond to idempotent elements of $End(M_R)$. 
Every once in a while, some structure in $M_R$ can tell you a lot about the structure of $End(M_R)$. For example, if $M_R$ is simple, $End(M_R)$ is a division ring; if $M_R$ is semisimple, then $End(M_R)$ is von Neumann regular; if the submodules of $M_R$ are linearly ordered, then $End(M_R)$ has either one or two maximal right ideals.
Those are just the ones off the top of my head... to be sure there are more.
