understanding of the "tensor product of vector spaces"
The first definition comes from the philosophy that students are bad at understanding abstract definitions and would prefer to see the tensor product defined as a space of functions of some kind. This is the reason that some books define the tensor product of $V$ and $W$ to simply be the space of bilinear functions $V \times W \to k$ ($k$ the underlying field), but this defines what in standard terminology is called the dual $(V \otimes W)^{\ast}$ of the tensor product.
For finite-dimensional vector spaces, $(V^{\ast})^{\ast}$ is canonically isomorphic to $V$, and that is the property that Gowers is taking advantage of in the first definition, which is basically a definition of $((V \otimes W)^{\ast})^{\ast} \cong V \otimes W$. The second definition is essentially the standard definition.
To answer your last question, no, we do not. $Z$ is infinite-dimensional whenever the underlying field is infinite. It is really, really huge, in fact pointlessly huge; the relations are there for a reason.
I'm going to go way out on a limb and instead of answering the questions actually posed, I'll propose a way to think about..... um OK, here it is: what's the difference between an ordered pair of vectors and a tensor product of two vectors? It is this: If you multiply one of the two vectors by $c$ and the other by $1/c$, then you've got a different ordered pair of vectors, but you've got the same tensor product. (Hence the tensor product is a quotient of the Cartesian product.)
As for your first question: yes.
As for your second question: for instance, elements of the basis of the "first" $V\otimes W$, make the expression
$$ [v, w+w'] - [v,w]- [v,w'] $$
to be equal to zero. Indeed, by definition, this guy evaluated on any bilinear map $f: V \times W \longrightarrow \mathbb{R}$ is
\begin{align} ([v, w+w'] - [v,w]- [v,w']) (f) &= [v, w+w'] (f) - [v,w] (f)- [v,w'] (f) \\ &= f(v,w+w') - f(v,w) - f(v,w') \\ &= 0 \end{align}
by the bilinearity of $f$.
In the same way, you can verify that elements of the basis of the "first" $V\otimes W$ make all the expression you're quotienting out in the second $V\otimes W$ to be zero. Hence, it is true that
$$ [v, w+w'] - [v,w]- [v,w'] = 0 $$
as well as
$$ [[v, w+w']] - [[v,w]]- [[v,w']] = 0 $$
in $Z/E$.
As for your third question: no.