I am having trouble understanding why ${B|}_{W_y} \not\equiv 0$.
Since I couldn't understand this part of the proof (on pg. 24 of the slides), I was trying to show it differently in my own way.
Assume ${B|}_{W_y} \equiv 0$.
Then, $B(ax+by,ax+by)=0$ for any $a,b \in \mathbb{F}$.
We can expand the equation above, and the expanded equation is $$2abB(x,y)+b^2 B(y,y)=0$$ since $B(x,x)=0$.
The equation implies that $b=0$ or $2aB(x,y)+bB(y,y)=0$.
However, $b =0$ is a contradiction since there exists vector in $W_y$ s.t $b \neq 0$.
Thus, $2aB(x,y)+bB(y,y)=0$ for any $a,b \in \mathbb{F}$.
And, this implies that $B(x,y)=B(y,y)=0$.
But then, $L$ is the conic by the definition of the conic $C$.
However, $x \notin L$, therefore it is also a contradiction.
Hence, ${B|}_{W_y} \not\equiv 0$.
This is significantly lengthier than the original proof in the link, but it's easier for me because I'm not good at linear algebra.
Is my answer OK?