How does one phrase the construction of the tensor product categorically? Given $U,V \in Vect$ we can build the tensor product as $U \otimes V := F(U \times V) / F(N)$ where $F:Set \rightarrow Vect$ is the free functor and where

$N:={(au,v) - a(u,v), (u,av) - a(u,v), (u+u',w) - (u,w) - (u',w), (u,v+v') - (u,v) - (u,v')  : a \in K; u, u' \in U; v, v' \in V }$

Then we have  the projection $p:U \times V \rightarrow U \otimes V$. The universalism of the projection, says that it is initial amongst all morphisms whose kernel includes $F(N)$. This means that these morphisms will also be bilinear. Thus we see that the tensor morphism is initial amongst all morphisms that are bilinear. 

How does one phrase this in the terminology of universal morphisms? The Wikipedia page gives an example of the tensor algebra where they start with a forgetful functor $U:K-Alg \rightarrow K-Vect$. 

I don't see how something similar can work here since the tensor product is an element of $Vect$. Yet it seems to me that the notion of a universal morphism here is appropriate.
The solution relies upon the fact that the tensor product is the coproduct in the category of commutative algebras. A proof of this is here in an answer in Math.SE by Omar Antolin-Carmenera. 

Then we need only note that the categorical product has a formulation as a universal morphism, in fact, a terminal morphism as pointed out in the second example on the Wikipedia article on universal morphisms (note that there is a mistake in the detail of the proof, though the idea is the right one).

Finally, we just need take the dual of this construction to formulate the coproduct as an initial morphism.
The solution relies upon the fact that the tensor product is the coproduct in the category of commutative algebras. A proof of this is here in an answer in Math.SE by Omar Antolin-Carmenera. 

Then we need only note that the categorical product has a formulation as a universal morphism, in fact, a terminal morphism as pointed out in the second example on the Wikipedia article on universal morphisms (note that there is a mistake in the detail of the proof, though the idea is the right one).

Finally, we just need take the dual of this construction to formulate the coproduct as an initial morphism.