When defining the tensor product $V \otimes W$, we define this as the quotient space $Free(V \times W) / \mathcal{R}$ where $\mathcal{R} \subset Free(V \times W) $ is an appropiately chosen subspace of the free vector space which makes the mapping $ v \otimes w := \mu(v,w) := \pi_{Free(V \times W) / \mathcal{R}}((v,w))$ bilinear.
Why do we require the quotient space $\mathbb{R}\langle V \times W \rangle / \mathcal{R} := V \otimes W$ rather than just directly using the quotient space $(V \times W) / \mathcal{R}$? We would have $V \otimes W := (V \times W) / \mathcal{R}$, which is a vector space and we could then define a bilinear mapping $\mu(v,w) := v \otimes w$ in the same way as above.
Thanks.
Edit (Before I wrote everything wrong... twice)
Just look how the generators of $R$ would look like if $R\subset V\times W.$ If you want to repeat construction from $Free(V\times W)$ you would get that $R=V\times W$. In fact: $$(v_1,w)+(v_2,w)=(v_1+v_2,w+w)$$ for example. So one generator of $R$ equals $$(v_1,w)+(v_2,w)-(v_1+v_2,w)=(0,w).$$
Similarly we can recoevr every $(v,0).$ hence $R=V\times W.$
Notice that if $V,W$ have finite dimensions, then the space $Free(V\times W)$ is huge and $R$ is also huge. However $V\otimes W:=Free(V\times W)/R$ has dimension $\dim V\cdot\dim W.$
I guess you have problems with understanding this construction cause you do not have enought intuition of what $Free(V\times W)$ is. I reccomend you to convince yourself that $Free(V\times W)$ is quite a big space (hint: look at its base).