I am reading the book of "Bland - Rings and their Modules (2011)".
Note: $R$ is not (necessary) commutative.
I want to take you to example 1 of section 1.5 on page 33. $M$ and $N$ are right R-modules. Bland claims that $\mbox{Hom}_R(M,N)$ is a left R-module if we define $[r\bullet f](x) = f(xr)$, for $r\in R$ and $x\in M$. He proves that $[s\bullet (r\bullet f)](x)=[(sr)\bullet f](x)$ for $r, s\in R$ and $x\in M$ and takes the other conditions that prove that $\mbox{Hom}_R(M,N)$ a module for granted.
However, he omits to prove that if $f$ is an R-map, then $rf$ is an R-map, i.e., if $f\in \mbox{Hom}_R(M,N)$ then $[r\bullet f] \in \mbox{Hom}_R(M,N)$, for $r\in R$. So, given $[r\bullet f](x) = f(xr)$ and $f(xa)=f(x)a$ then it follows that $[r\bullet f](xa)=[r\bullet f](x)a$, for $r, a\in R$ and $x\in M$.
Who can give me a proof of this or is Bland wrong?
R is not commutative: I know that the text of this example is not very clear, but Bland treats the commutative case seperatly in this example, he also claims this for right and left R-modules, why do that if R is commutative?
The author is wrong, as you rightfully pointed out: there is no reason for the equality $$(rf)(xa) = f(xar) \stackrel{!}{=} f(xra) = f(xr)a = (rf)(x)a$$ to hold.
For an explicit counterexample, take $M = N = M_2(\mathbb{Z})$ as right-modules over themselves, $f = 1_M$, and $x = \begin{pmatrix}1&0\\0&1\end{pmatrix}$, $r = \begin{pmatrix}1&0\\0&0\end{pmatrix}$, $a = \begin{pmatrix}0&1\\0&0\end{pmatrix}$. Then $f(xar) = ar = 0$ and $f(xra) = ra = a \neq 0$.
For completeness, here are the proper statements (taken from Rotman's book):