My text is based on these notes:
http://outcomes.enquiringminds.org/definition-and-construction-of-the-tensor-product/
http://www-users.math.umn.edu/~broom010/doc/TensorProduct.pdf
There are one point that I'm really struggling to understand.
Why use the quotient of free vector space $F(V\times W)$ and $R(V\times W)$ ?
$R(V\times W)$ is spanned by these vectors:
\begin{equation} (v_{1}+v_{2},w) - (v_{1},w) - (v_{2},w) \end{equation}
\begin{equation} (v,w_{1}+w_{2}) - (v,w_{1}) - (v,w_{2}) \end{equation}
\begin{equation} (cv,w) - c(v,w) \end{equation}
\begin{equation} (v,cw) - c(v,w) \end{equation}
There is a natural map $\iota: V \times W \to F(V \times W)$ given by $(v, w) \mapsto (v, w)$. However, this map is not bilinear - and we're doing linear algebra, so we want our maps to be (bi)linear. So we want to add a bunch of relations into $F(V \times W)$ such that this map does become bilinear; and adding a bunch of relations into a vector space is precisely taking a quotient vector space.
We use this particular quotient because $R(V \times W)$ is the minimal subspace of $F(V \times W)$ such that the composition $V \times W \to F(V \times W) \to F(V \times W) / R(V \times W)$ is bilinear (where the second map is the natural projection map). That is, $R(V \times W)$ contains all the relations needed to make $\iota$ bilinear, and no more.