Why is the tensor product constructed in this way? I've already asked about the definition of tensor product here and now I understand the steps of the construction. I'm just in doubt about the motivation to construct it in that way. Well, if all that we want is to have tuples of vectors that behave linearly on addition and multiplication by scalar, couldn't we just take all vector spaces $L_1, L_2,\dots,L_p$, form their cartesian product $L_1\times L_2\times \cdots \times L_p$ and simply introduce operations analogous to that of $\mathbb{R}^n$ ? 
We would get a space of tuples of vectors on wich all those linear properties are obeyed. What's the reason/motivation to define the tensor product using the free vector space and that quotient to impose linearity ? Can someone point me out the motivation for that definition ?
Thanks very much in advance.

The best example to look at why the tensor product is defined as it is is by considering bilinear functions $B : V \times V \rightarrow \mathbb{R}$
Suppose we wish to represent $B$ by a single linear map of vector spaces $\hat{B} : X \rightarrow \mathbb{R}$.  What sort of basis would our space $X$ need to have in order to define $\hat{B}$ from $B$?
Let $\{e_i\}$ be a basis for $V$, then each term $B(e_i, e_j)$ can be defined independently.  The reason we can't simply pick $X$ to be $V \times V$ is that if we define $\hat{B}(x) = B(x_1, x_2)$, then $$\hat{B}(e_i, e_j) = \hat{B}(e_i, 0) + \hat{B}(0, e_j) = B(e_i, 0) + B(0, e_j) = 0 + 0 = 0$$
where $x = (x_1, x_2) \in V \times V$.
Instead we need $X$ to be generated by pairs of basis vectors themselves, $X = \langle e_i \otimes e_j\rangle $ and then we can think of $\hat{B}$ as a linear operator on this space.
