Suppose a have two $R$-modules, $M$ and $N$ and suppose that $M$ has basis $E$. I want to know if we can always obtain an $R$-module homomorphism just by specifying values on $E$.
More formally does any function $f : E \to N$ induce an $R$-module homomorphism $\tilde{f} : M \to N$?
I thought that since for each $x \in M$ we have $$x = \sum_{e \in E} r_ee$$ we could then define $$\tilde{f}(x) = \sum_{e \in E} r_e f(e).$$ However I'm not sure if $\tilde{f}$ is well-defined (since I'm not sure if every element of $M$ is can be written uniquely in terms of basis elements).
As others have said in the comments $\bar{f}$ is well defined since $E$ is linear independent and therefore your decomposition of $x$ is unique. It's also pretty obvious that $\bar f$ is $R$-linear.
This does not work for non-free modules, i. e. modules that lack a basis. For Example take the $\mathbb{Z}$-Module $\mathbb{Z}/p\mathbb{Z}$ for $p$ prime. It is generated by any of its elements except the equivalence class $\bar 0$ of $0$. But for every $a \in \mathbb{Z}/p\mathbb{Z} \setminus \{\bar 0\}$ there is no linear mapping $\mathbb{Z}/p\mathbb{Z} \to \mathbb{Z}$ extending $a \mapsto 1$.