I've only seen two ways of seeing the tensor product: (1) for vector spaces, where one uses a basis, and (2) the universal property, that describes the point of the tensor product. This is the first hands-on-but-not-basis construction of the tensor product I've seen, and so I record it here. Let $R$ be the base ring. We are tensoring the modules $M$ and $N$.

• Create the free $R$ module $F \equiv R^{M \times N}$ whose elements are free linear combinations of tuples from $M$ and $N$.
• Let $K$ (for kill) be the submodule generated by the equations: (1) $(x + x', y) - (x, y) - (x', y)$, (2) $(x, y + y') - (x, y) - (x, y')$, (3) $(rx, y) - r \cdot (x, y)$(4) $(x, ry) - r \cdot (x, y)$
• Let $T \equiv F/K$. Denote the element of each $(x, y) \in F$ in the quotient $F/K$ as $x \otimes y$. Then $T$ is generated by such elements.
• From our quotients, we have the equation $(ax) \otimes y = a (x \otimes y) = x \otimes (ay)$, and $(x + x') \otimes y = x \otimes y + x' \otimes y$, and finally $x \otimes (y + y') = x \otimes y + x \otimes y'$.

Any $R$-module map map $f: M \times N \rightarrow O$ where $O$ is an $R$-module extends by linearity into $f_F: F \rightarrow O$ as $f_F(\sum_i r_i (x, y)) \equiv r_i f(x, y)$. We also have a map $-/\sim : F \rightarrow T$

 F=R^{MxN} →fF→ MxN →f→ O ↓               ↑ F/~             fT ↓               ↑ T=M(x)N →→fT→→→→*

For this diagram to commute, we need the fibers of $(f \circ f_F): R^{MxN} \rightarrow O$ to take constant values for each $o \in O$. Unwrapping that condition implies that $f$ is bilinear. So, the condition $f_T(x \otimes y) = f(x, y)$ uniquely determines $f_T(x \otimes y)$ if $f(x, y)$ is bilinear. If not, the map $f_T$ is ill-defined, as we cannot "kan extend" $f_F$ along $f_T$.