I am attempting to prove that $\operatorname{Hom}_A(M,A)$ is free given that $M$ is a free $A$ module.
So, currently I am using the fact that $\phi \in \operatorname{Hom}_A(M,A)$ is dependent on how it acts on our basis $\{e_1,...,e_n\}$ for our $A$ module $M$. Thus by defining our basis of homomorphism as $\{\phi_1,...,\phi_n\}$ and $\phi_i(e_j)=\delta_{i,j}$.
However, we will need to give some specificity to $\delta_{i,j}$, to be honest I am very confused. Maybe the fact my $A$ module $M$ is module isomorphic to $A^n$ may help...
I would appreciate a push in the right direction. Sorry about the vagueness of my attempt.
Assume that $A$ is a commutative ring with unity $1_A$. Then it works exactly as in the vector space: $\delta_{i,j}=1_A$ if $i=j$, otherwise $\delta_{i,j}=0_A$, where $0_A$ is the identity element of the additive abelian group underlying the ring $A$.
Actually, in this way we can show that there is an isomorphism of free modules (of finite rank) over $A$: $$M\cong \text{Hom}_A(M,A),$$ defined exactly as you suggested on a basis, and then extended to all elements.