Well-definedness of tensor product of modules Given two $R$-modules $M$ and $N$, one possible construction of the tensor product $M\otimes_R N$ is by taking the quotient $F/U$ of the free module $F=\bigoplus_{m,n}R\cdot(m,n)$ and the submodule $U$ of $F$ generated by elements of the form $a(m,n)-(am,n)$ etc. The corresponding bilinear map $\tau\colon M\times N\to M\otimes_R N$ then simply is the projection $(m,n)\mapsto \pi_U(m,n)$.
In order to show that this actually defines a tensor product, one has to prove that the universal property holds: Given a bilinear map $\psi\colon M\times N\to L$, there is a unique linear map $\hat\psi\colon M\otimes_R N\to L$ such that $\hat\psi\circ \tau=\psi$. The uniqueness is obvious because the previous condition leads to $\hat\psi\overline{(m,n)}=\psi(m,n)$, which extends to whole $M\otimes_R N$ by linearity; thus, uniquely determining all of the values of $\hat\psi$. What I am struggling to understand is why this map is also well-defined. There could be two different representations $\sum_{m,n} a_{m,n} \overline{(m,n)}=\sum_{m,n} a_{m,n}' \overline{(m,n)}$ of the same element in $M\otimes_R N$, which could get mapped to different values under $\hat\psi$, namely to $\sum_{m,n} a_{m,n} \psi(m,n)$ and $\sum_{m,n} a_{m,n}' \psi(m,n)$ in $L$.
How to show that $\hat\psi$ is actually well-defined?
 A: actual answer:
The key to the well-definedness of $\hat\psi$ is that $\psi : M \times N \to L$ is a $R$-bilinear mapping, so, if we extend it by linearity on $F$, it vanishes on $U$ (it vanishes on each one of its generators because of $R$-bilinearity), so the map $\hat\psi : F/U \to L$ where $\hat\psi(\overline{x}) = \psi(x)$ for any $x \in F/U$ must be well defined, since
$\overline{x} = \overline{y} \implies x-y \in U \implies \hat\psi(\overline{x-y})=0 \implies \hat\psi(\overline{x}) = \hat\psi(\overline{y})$
previous answer (before edit in the question):
Simply put, your statement that there could me two different representations $\sum\limits_{m,n}a_{m,n}(m,n)$, $\sum\limits_{m,n}a'_{m,n}(m,n)$ is wrong, since you're not looking at $M \times N$ as a ring or module, you're looking at the set $F$, which is the free $R$-module that has $M \times N$ as a basis, they're very different in nature.
So we have that
$\sum\limits_{(m,n) \in M \times N}a_{m,n}(m,n) = \sum\limits_{(m,n) \in M \times N}a'_{m,n}(m,n) \implies \sum\limits_{(m,n) \in M \times N}(a_{m,n}-a'_{m,n})(m,n) = 0$
and from that, since its a basis, we must have that $a_{m,n} - a'_{m,n} = 0 \implies a_{m,n} = a'_{m,n}$ for all $(m,n) \in M \times N$.
