Let $A$ be a ring with $1$ and $M$ a left $A$-module. In chapter 6 of Curtis and Reiner's 'Methods of Representation theory', $M$ is regarded as a right module over the endomorphism ring $E=\text{End}_A(M)$ of $M$.
I'm confused how $M$ can become a right $E$-module as the only $E$-action on $M$ I can think of is by $E$ acting as left operators on $M$ making it a left $E$-module.
I must be overlooking something trivial here.
On
To obtain a right module $M_E$ we (implicitly) have to switch the direction of the operation of $E$, i.e. have to interpret composition of endomorphisms from left to right. This way, with the original left action ${}_AM$, we get a bimodule ${}_AM_E$ as the connecting axiom $$(am)\varphi=a(m\varphi)$$ follows from $\varphi$ being an $A$-module homomorphism.
I don't know the book by Curtis and Reiner, but a well-established convention in ring and module theory is to write morphisms on the opposite side to the scalars and probably the book uses it.
This avoids awkward references to the opposite ring that pop out when morphisms are written on the same side.
So, if left module are considered, morphisms are written on the right and composition of maps reflects this: so, for $f\colon L\to M$ and $g\colon M\to N$, the composition is $fg$ $$ (x)fg=((x)f)g $$ for $x\in L$.
This has the consequence that a left $A$-module is in a natural way a right module over the endomorphism ring $B=\operatorname{End}({}_AM)$ of $_AM$.
The natural map $A\to\operatorname{End}(M_B)$ is a ring homomorphism; it is defined by $a\mapsto \hat{a}$, where $\hat{a}(x)=ax$. This is indeed an endomorphism of $M_B$, because, for $b\in B$, $$ \hat{a}(xb)=a((x)b)=(ax)b=(\hat{a}(x))b $$ (the middle equality is because $b\in\operatorname{End}({}_AM)$).
In particular, $M$ becomes an $A$-$B$-bimodule.