Let $H$ be a Hopf algebra and $M, N$ two left $H$-modules, then we can define a left $H$-module structure on $\mathrm{Hom}(M,N)$ given by $(h * f)(m) = h_1 f(S(h_2) m)$, where $S$ is the antipode. My question is kind of further from that, I want to show that $*$ acts on $\mathrm{Hom}(M, M)$ such that $$ h * (f \circ g) = (h_1 * f) \circ (h_2 * g) \quad\text{and}\quad h * \mathrm{id}_M = \epsilon(h) \mathrm{id}_M \,. $$
The second identity is easier to show: given $m \in M$, we have $$ (h * \mathrm{id}_M)(m) = h_1 \mathrm{id}_M(S(h_2) m) = h_1 S(h_2) m \,, $$ but $h_1 S(h_2) = \epsilon(h)$, so $(h * \mathrm{id}_M)(m) = \epsilon(h) m = \epsilon(h) \mathrm{id}_M(m)$.
Now I'm stuck on $h * (f \circ g) = (h_1 * f) \circ (h_2 * g)$, because for each $m \in M$ the left side is $$ (h * (f \circ g))(m) = h_1 f(g(S(h_2) m)) $$ and for the other side we have $$ ((h_1 * f ) \circ (h_2 * g))(m) = (h_1 * f)(h_{2,1} g(S(h_{2,2}) m)) = h_{1,1}f(S(h_{1,2})h_{2,1}g(S(h_{2,2})m)) \,. $$
I don't know how to rewrite that expression to arrive at some equality, can you help me?
If I'm interpreting your notation correctly, then what you need is to apply coassociativity. Using Sweedler notation, this means that your $h_{1,1} \otimes h_{1,2}\otimes h_{2,1}\otimes h_{2,2}$ can really be reindexed as $h_{(1)}\otimes h_{(2)}\otimes h_{(3)}\otimes h_{(4)}$, because $(\Delta\otimes\operatorname{id}) \circ \Delta = (\operatorname{id}\otimes \Delta)\circ \Delta$ and so on.
The identity you want is $h \ast (f\circ g) = (h_{(1)}\ast f) \circ (h_{(2)}\ast g)$. You have expanded the right hand side using the definition of the $H$ action and got $ (h_{(1)}\ast f) \circ (h_{(2)}\ast g)(m) = h_{(1)} f(S(h_{(2)}) h_{(3)} g(S(h_{(4)})m))$. The factors $S(h_{(2)}) h_{(3)}$ can be collapsed using $S(h_{(1)})h_{(2)} = \epsilon(h)$, which can then be pulled out of $f$, so we can simplify the result to $h_{(1)} \epsilon(h_{(2)}) f(g(S(h_{(3)}m)) = h_{(1)} f(g(S(h_{2})m)) = h\ast(f\circ g) (m)$.