For each pair $(V,W)$ of vector spaces (over a fixed ground field), let $T(V,W)$ be their tensor product, and $F(V,W)$ the vector space of bilinear forms on $V^*\times W^*$. One checks that
(a) there is a unique linear map $e(V,W)$ from $T(V,W)$ to $F(V,W)$ satisfying
$$
\big(e(V,W)(v\otimes w)\big)(f,g)=f(v)g(w)
$$
for all $v\in V,w\in W,f\in V^*,g\in W^*$,
(b) $e(V,W)$ is injective,
(c) $T$ and $F$ are functors,
(d) $e$ is a natural transformation from $T$ to $F$,
(e) $T,F$ and $e$ are compatible (in an obvious sense) with finite direct sums.
Claim 1: $e(V,W)$ is surjective $\iff$ the cardinal number $\dim(V)\dim(W)$ is finite.
In view of (b), implication "$\Leftarrow$" follows by dimension counting. It suffices thus to prove the non-surjectivity when $V$ is infinite dimensional and $W$ nonzero. Writing $W$ as $W_1\oplus W_2$ with $\dim W_1=1$ and using (b), we are reduced to
Claim 2: if $V$ is infinite dimensional, then the canonical embedding $V\to V^{**}$ is not surjective.
To prove this, we'll use an embedding of $V$ in $V^*$, and an embedding of $V^*$ in $ V^{**}$. None of these two embeddings will be canonical, but their composition will.
Choose a basis $B$ of $V$, and identify $V$ to the space $K^{(B)}$ of finitely supported $K$-valued functions on $B$. Then $V^*$ can be identified to the space $K^B$ of all $K$-valued functions on $B$. Similarly, we can identify $V^*$ to $K^{(B\sqcup C)}$, where $C$ is a set and $\sqcup$ means "disjoint union". As $B$ is infinite, $C$ is nonempty. Using the same trick once more, we can identify $V^{**}$ to $K^{(B\sqcup C\sqcup D)}$, where $D$ is a nonempty set. Then the natural embedding of $K^{(B)}$ in $K^{(B\sqcup C\sqcup D)}$, which is clearly not surjective, corresponds to the natural embedding of $V$ in $V^{**}$. This completes the proof.
One technique that is useful in algebra is to try to relate the subspace you are working with (in this case $V_1 \times_W V_2$) with some linear map.
Define a map $C: V_1 \times V_2 \to W$ where $C((v_1, v_2)) = Av_1 - Bv_2$. Note that $C$ is linear, $\ker C = V_1 \times_W V_2$, and $\operatorname{range} C = \operatorname{range}A + \operatorname{range}B$.
By the Rank-Nullity Theorem $$\dim(V_1 \times V_2) = \dim(\ker C) + \dim(\operatorname{range} C)$$ so $$\dim(V_1) + \dim(V_2) = \dim(V_1 \times _W V_2) + \dim(\operatorname{range}A + \operatorname{range}B)$$ and the desired equality follows.
Best Answer
These are very different. Are you aware of what the free vector space on a set $X$ is? It is a vector space $\mathscr F(X)$ with $X$ as a basis. In particular, if you choose to be $X$ to be equal to be a real vector space $V$, then $\mathscr F(V)$ is a real vector space which has $V$ as a basis. So if, for example, $V=\Bbb R$, then $V$ is a finite dimensional real vector space, but $\mathscr F(V)$ is a vector space with dimension of cardinality $\Bbb R$, and $\mathscr F(V)$ is indeed very different to $V$.
To say it in a different way, $\mathscr F(V)$ is all "formal combinations" of elements of $V$. The typical addition and scalar multiplation from $V$ doesn't apply here. So in particular, in $\mathscr F(V\times V')$, the elements $(v,0)+(0,v')$ and $(v,v')$ are different elements, even though they are the same as elements of $V\times V'$.