$ \newcommand{\Hom}{\mathrm{Hom}} $
If $i:A\to A$ and $j:B\to B$ are the identity homomorphisms of the modules $A$ and $B$ over $R$, then:
$$\Hom(i,j):\Hom(A,B)\to\ Hom(A,B)$$
is the identity homomorphism of the module $\Hom(A,B)$. If $f:A'\to A$, $f':A''\to A'$, $g:B\to B', g':B'\to B''$ are homomorphisms of modules over $R$ we have
$$\Hom(f\circ f', g'\circ g) = \Hom(f', g')\circ \Hom(f,g)$$
Just as a clarification, my book defines $\Hom(f,g)$ as being the following thing:
Let $f:A'\to A$ and $g:B\to B'$ denote arbitrarily given homomorphisms of modules over $R$, and consider the modules $\Hom(A,B)$ and $\Hom(A', B')$. Define a function
$$h:\Hom(A,B)\to \Hom(A', B')$$
by taking $h(\phi) = g\circ \phi\circ f$
for every $\phi\in \Hom(A,B)$. Clearly $h$ is a homomorphisms of the module $\Hom(A,B)$ into the module $\Hom(A', B')$ which will be denoted by the symbol
$$\Hom(f,g)$$
In order to prove $\Hom(f\circ f', g'\circ g) = \Hom(f', g')\circ \Hom(f,g)$, I must see $f\circ f'$ as one function, and $g'circ g$ as another, and apply the definition given in the book. So my $h$ will be a function such that $h(\phi) = (g'\circ g)\circ \phi \circ (f\circ f')$. I tried to use associativity, for example, to see $g'\circ (g\circ \phi \circ f)\circ f'$. I think I can kinda see $\Hom(f,g)$ in the middle, and $\Hom(f',g')$ in the points, but I'm lost. Could somebody help me?
$ \newcommand{\Hom}{\mathrm{Hom}} $ You almost have the solution there. Let us give the functions names to make things a bit easier on ourselves. \begin{equation} h_1 = \Hom(f \circ f',g' \circ g), \, h_2= \Hom(f',g'), \, \text{ and }\, h_3 = \Hom(f,g) \end{equation} Then by definition, \begin{equation}h_1(\phi)=g' \circ g \circ \phi \circ f\circ f' \end{equation} Likewise we have, \begin{equation}h_2 \circ h_3(\phi)=h_2(g \circ\phi \circ f)= g' \circ g \circ \phi \circ f \circ f' \end{equation} So we have \begin{equation}\Hom(f \circ f',g' \circ g) = h_1 = h_2 \circ h_3 = \Hom(f',g') \circ \Hom(f,g) \end{equation} as desired.