Set of Homomorphisms as an $R-$ module $\require{AMScd}$
I'm reading A first course of homogocial algebra by D.G. Northcott, and I don't quite get the Example 1 on page 25. Here's what it says:
Example 1

Let the module $A$ belongs to $\mathscr{C}_R$ (i.e the category of $R-$modules). Since $R$ is an $(R; R)-$bimodule. $\mbox{Hom}(R; A)$ is an $R-$module of the same type as $A$.

The bolded part is where I don't get it.
So, say I have $A$ as a left $R-$module, now I'll try to define the outer multiplication of $\mbox{Hom}(R; A)$, so to make it also a left $R-$module.
From this part on, I'll consider $f \in \mbox{Hom}(R; A)$; $r, \alpha \in R$.
Trial 1
I define $(rf)(\alpha) = f(r\alpha)$. I don't think this should work, as:
$\begin{CD}
(r_1f)(r_2\alpha) @= r_2(r_1f)(\alpha) @= r_2f(r_1\alpha) @= \color{red}{r_2r_1}f(\alpha) \\ @| \\ f(r_1r_2\alpha) \\ @| \\ \color{red}{r_1r_2}f(\alpha)
\end{CD}$
Trial 2
I define $(rf)(\alpha) = f(\alpha r)$. I don't think this should work, either:
$\begin{CD}
(r_1r_2f)(\alpha) @= (r_1(r_2f))(\alpha) @= r_1f(\alpha r_2) @= f(\alpha \color{red}{r_2 r_1}) \\ @| \\ f(\alpha \color{red}{r_1r_2})
\end{CD}$

I think I'm missing something really big here. I would be so glad if you guys can help me.
Thanks a lot in advance,
And have a good day,
 A: It might be visually helpful to track the module side in the notation. To do that, I'll write $A_R$ to indicate a right $R$ module $A$, or $_RA$ to indicate a left $R$ module.
It will also help to consider bimodules over two different rings, since it will help keep us from confusing the module operations.
Suppose $_SB_R$ is a bimodule.
$H=Hom(B_R,A_R)$ has a natural right $S$ module structure via $(f\cdot s)(x):=f(sx)$.
Thus if $r$ is another element of $R$, $(f\cdot s)(xr)=f(sxr)=f(sx)r=(f\cdot s)(x)r$. You can see that the module actions are taking place on opposite sides of the ring, and so they don't need to "cross over" each other.
If $_TA_R$ is a bimodule, then $H$ has a natural left $T$ module structure given by
$(t\cdot f)(x):=tf(x)$. Again, $(t\cdot f)(xr)=tf(xr)=tf(x)r=(t\cdot f)(x)r$. The same thing has happened here that the $R$ module action and $T$ module action do not have to cross each other.
Both can happen at once: If $_TA_R$ and $_SB_R$ are bimodules, then $_TH_S$ is a bimodule, using both operations above (the compatibility $(th)s=t(hs)$ has to be checked.)
The problems with your definitions are that you are crowding the side of modules in the Hom and the new module action on one side, where they need to "cross over" each other in order to come out right. They really should be taking place on opposite sides of each other, as with these two definitions.

Let's work out the "left hand version" of this, which it looks like you'll need, since you're considering $H=Hom(_RR,_RA)$. The natural left module structure would arise from the right module structure of $R_R$. That is, $(r\cdot f)(x):=f(xr)$ would be a left $R$ linear map: $(r\cdot f)(r_1x)=f(r_1xr)=x_1(f(xr))=x_1(r\cdot f)(x)$
