Why $(f\mapsto f(v_i)w)_{i,j}$ with $f\in V'$,$w\in W$ is a basis of $\mathscr{L}(V',W)$? I'm trying understand the proof of the Proposition 3.1.2 (pg.5) of this document: http://www.win.tue.nl/~amc/ow/lba/lba3.pdf

Suppose $V$ and $W$ are finite dimensional. If $(v_i)_i$ is a basis of $V$ and $(w_j)_j$ is a basis of $W$, then $(v_i\otimes w_j)_{ij}$ is a basis of $V\otimes W$

Why $(f\mapsto f(v_i)w_j)_{i,j}$ with $f\in V'$,$w\in W$ is a basis of $\mathscr{L}(V',W)$?
My attempt to understand:
Let $g\in\mathscr{L}(V',W)$. So $g(f)=w=\sum\limits_j^{dim(W)}w_jx_j$. So $g=(f\mapsto w=\sum\limits_j^{dim(W)}w_jx_j)$. 
My ideas:
We have that $\exists F\in V'$ such that $F(v_j)=x_j$. $\checkmark$ 
Why does the author say that $h = F$ for arbitrary $h\in V'$? 
If $(f\mapsto f(v_i)w)_{i,j}$ is a base of $\mathscr{L}(V',W)$ how to write a vector in this basis?
 A: It seems you want to be able to work this out for concrete vectors, so I will try to give examples. You will be much more effective in working with tensor products, if you also learn to work with the abstract properties.
Following the notes, theorem 3.1.2 gives an isomorphism 
$$V\otimes W \xrightarrow{k} \mathscr{L}(V',W),$$
$$v \otimes w \mapsto (f \mapsto f(v)w).$$
This is the key part of the proof. This is proved using the universal property of tensor products.
Note that a basis of $V\otimes W$ is $(v_i\otimes w_j )_{i,j}$ with $(v_i)_i$ a basis of $V$ and similarly for $(w_j)_j$ and $W$. It is an easy exercise to work this out.
Combine these two observations. Under an isomorphism a basis is send to a basis, so $k((v_i\otimes w_j)_{i,j} )$ is a basis for $\mathscr{L}(V',W)$.
Using our map $k$ $$k((v_i\otimes w_j )_{i,j})= (f \mapsto f(v_i)w_j)_{i,j}.$$
So that is why $(f \mapsto f(v_i)w_j)_{i,j}$ is basis for $\mathscr{L}(V',W)$.
If you want to know how to express an arbitrary elemen $y\in \mathscr{L}(V',W)$ in this basis, you can figure out the inverse to $k$ and use the more intuitive basis of $V\otimes W $.
