If $M$ is a right $R$-module, then $\operatorname{Hom}(R \times R,M)$ is isomorphic to $\operatorname{Hom}(R,M) \times \operatorname{Hom}(R,M)$
My questions are:
What is the isomorphism map? (I tried this $f\to(f_1,f_2)$ where $f_1(r) =f(r,0)$ and $f_2(r)=f(0,r)$. Is it correct?)
Can we make this natural group isomorphism a right $D$-module isomorphism, where $D=\operatorname{End}(R\times R)$ is set of all right $R$-homomorphisms from $R \times R$ to $R\times R$?
(1) More generally, if $X$ and $Y$ are $R$-modules, you have $$ \operatorname{Hom}(X\oplus Y,M)\cong \operatorname{Hom}(X,M)\oplus\operatorname{Hom}(Y,M) $$ (in module theory it's more customary $\oplus$ rather than $\times$, but they mean the same).
The (group) isomorphism is obtained by considering the canonical injections $i\colon X\to X\oplus Y$ and $j\colon Y\to X\oplus Y$ defined by $$ i(x)=(x,0),\quad j(y)=(0,y) $$ and sending $f\colon X\oplus Y\to M$ to the pair $(f\circ i,f\circ j)$, which is exactly what you did.
(2) Hint: the endomorphism ring of $R\oplus R$ is the ring of $2\times 2$ matrices over $R$. What would be the action?
