Let $A$ be a ring, $E$ a right $A$-module and $F$ a left $A$-module. Let $(b_\mu)_{\mu\in M}$ be a basis of $F$. Then every element of $E\otimes_AF$ can be written uniquely in the form $\sum_{\mu\in M}(x_\mu\otimes b_\mu)$ where $x\in E^{(M)}$.
Attempt:
The mapping $v:\bigoplus_{\mu\in M}Ab_\mu\rightarrow F,\,y\mapsto\sum_{\mu\in M} y_\mu$ is an $A$-module isomorphism. Furthermore, there exits a $\mathbf{Z}$-linear bijection $$g:E\otimes_A\bigoplus_{\mu\in M}Ab_\mu\rightarrow\bigoplus_{\mu\in M}(E\otimes_AAb_\mu)$$ such that $g(x\otimes(y_\mu)_{\mu\in M})=(x\otimes y_\mu)_{\mu\in m}$ for $x\in E$ and $y\in\bigoplus_{\mu\in M}Ab_\mu$. Thus the mapping $$[1_E\otimes v]\circ g^{-1}:\bigoplus_{\mu\in M}(E\otimes_AAb_\mu)\rightarrow E\otimes_A F$$ is a $\mathbf{Z}$-isomorphism. At this point, I don't how to deduce the required property: that every element $z\in E\otimes_A F$ can be uniquely written in the form $\sum_{\mu\in M}(x_\mu\otimes b_\mu)$ where $x\in E^{(M)}$. Any suggestions?
Start with a pure tensor $x\otimes y$, with $x\in E$ and $y\in F$. By assumption, $$ y=\sum_{\mu\in M}a_\mu b_\mu $$ and therefore $$ x\otimes y=\sum_{\mu\in M}(xa_\mu)\otimes b_\mu $$ Now passing to generic elements of $E\otimes_A F$ is easy.
What about uniqueness? Consider the isomorphism $g\colon A^{(M)}\to F$, $g((a_\mu)_\mu)=\sum_\mu a_\mu b_\mu$ Then define $$ \tau\colon E\times F\to E^{(M)},\qquad \tau(x,g((a_\mu)_\mu))=(xa_\mu)_\mu $$ Prove it is balanced, so it induces a group homomorphism $E\otimes_A F\to E^{(M)}$ and show that an obvious map is its inverse.