Decomposition of Bilinear Form Let $E=\text{span}\{x_{1},...,x_{n}\}$ and $F=\text{span}\{y_{1},...,y_{n}\}$. Assume that $B$ is a bilinear form on $E\times F$, the author claims that one can write 
\begin{align*}
B(x,y)=\sum_{j=1}^{m}\theta_{j}(x)\omega_{j}(y),
\end{align*}
where $\theta_{j}\in E^{\#}$, $\omega_{j}\in F^{\#}$, the (algebraic) linear forms.
I thought at the very beginning that the linear forms must be of the kind that $\varphi_{i}(e_{k})=\delta_{k,i}$ and $\psi_{j}(f_{k})=\delta_{k,j}$ for bases $\{e_{i}\}_{i=1}^{N}$ for $E$, and $\{f_{j}\}_{j=1}^{M}$ for $F$, but this does not give me any further to proceed to, as 
\begin{align*}
B(x,y)&=B\left(\sum_{i=1}^{N}\lambda_{i}e_{i},\sum_{j=1}^{M}\nu_{j}f_{j}\right)\\
&=\sum_{i=1}^{N}\sum_{j=1}^{M}\lambda_{i}\nu_{j}B(e_{i},f_{j}),
\end{align*}
the crossed term $B(e_{i},f_{j})$ cannot be reduced, I wonder some magic must be going on, so how to get the author's claim?

Consider the following : for each $(i,j)$, put $\theta_{(i,j)} = B(e_i, f_j) e_i^*$ (where $e_i^*(\sum_k \lambda_k e_k ) = \lambda_i$) and $\omega_{(i,j)} = f_j^*$. 
Then compute $\sum_{(i,j)} \theta_{(i,j)}\omega_{(i,j)} (x,y) = \sum_i \sum_j e_i^*(x)f_j^*(y)B(e_i,f_j)$, so with your notations,  $\sum_{(i,j)} \theta_{(i,j)}\omega_{(i,j)} (x,y) = \sum_i \sum_j \lambda_i \nu_j B(e_i,f_j) = B(x,y)$.
So $B= \sum_{(i,j)} \theta_{(i,j)}\omega_{(i,j)}$.
Note that you could have also chosen something else, such as $\theta'_{(i,j)} = e_i^*, \omega'_{(i,j)} = B(e_i,f_j)f_j^*$, or still other choices. 
The underlying idea is that $(E\otimes F)^* \cong E^* \otimes F^*$ for finite dimensional vector spaces $E,F$
