Let $A$ be a Commutative ring and $R$ be a commutative $A$-algebra. So every $R$-module has a natural $A$-module structure, and for every $A$-module $W$, and $R$-module $M$, the $A$-module $\text{Hom}_A(W,M)$ also has a compatible $R$-module structure defined as $r\cdot f(w):=r.f(w), \forall w\in W$, where $f\in \text{Hom}_A(W,M)$.
Let $U,V$ be $R$-modules and $f: U \to V$ be an $R$-module homomorphism and $W$ be an $A$-module. Then we have an $R$-linear map
$$\text{Hom}_A(W,f): \text{Hom}_A(W,U) \xrightarrow{g \mapsto f \circ g} \text{Hom}_A(W,V) $$ .
My question is: Do we have the following isomorphism of $R$-modules:
(1) $\ker \text{Hom}_A(W,f) \cong \text{Hom}_A(W,\ker f) $ ?
(2) $co\ker \text{Hom}_A(W,f) \cong \text{Hom}_A(W, co\ker f) $ ?
(1) Yes, $g\in\ker(\operatorname{Hom}_A(W,f))$ iff $f\circ g=0$ iff $\operatorname{im}(g)\subseteq\ker(f)$ iff $g\in\operatorname{Hom}_A(W,\ker(f))$.
(2) No,take $A=R=U=\mathbb{Z}$, $V=W=\mathbb{Z}/2\mathbb{Z}$ and $f\colon\mathbb{Z}\rightarrow\mathbb{Z}/2\mathbb{Z}$ to be the projection. Then $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z})=0$ and $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}$, so that $\operatorname{coker}(\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},f))=\mathbb{Z}/2\mathbb{Z}$. On the other hand, $\operatorname{coker}(f)=0$, so that $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},0)=0$.
The point in $(2)$ is the following. Assume $f$ surjective for simplicity, then $\operatorname{coker}(f)=0$ and $\operatorname{Hom}_A(W,\operatorname{coker}(f))=0$, so that your isomorphism postulates $\operatorname{coker}(\operatorname{Hom}_A(W,f))=0$, i.e. that $\operatorname{Hom}_A(W,f)$ is surjective. Spelled out, this would mean that for any surjective $f\colon U\rightarrow V$, any map $W\rightarrow V$ lifts to a map $W\rightarrow U$ through $f$. Such lifting can in general not be done in general, e.g. for $\operatorname{id}_V$ (as in the example), it already implies the existence of a right-inverse to $f$.
Also, let me stress that none of this has anything to do with putting both $A$- and $R$-module structures on things. It is all true/false on the level of $A$-modules for fixed $A$ already and then putting compatible $R$-module structures on the objects post hoc doesn't change anything.