Show that $\widetilde{f}$ is a nondegenerate map? Noted by $V, W$ two vectors spaces over the same field $K$ of finite dimensions. Let $f:V\times V\rightarrow W$ a degenerate map. I would show that
$$\widetilde{f}:\widetilde{V}\times \widetilde{V}\rightarrow W$$
is a nondegenerate map, where $\widetilde{V}:= V/\ker{f}$, with $\ker{f}=\{x\in V: \, f(x,y)=0\quad  \forall y\in V\}$ is the subspace kernel of $f$.
Recall that:
. A degenerate bilinear form $f:V\times V\rightarrow W$  on a finite-dimensional  vector space $V$ is a bilinear form such that it has a non-trivial kernel. And it is nondegenerate iff it's subspace kernel is trivial.
Thank you in advance
 A: This is a good question to see what happens in passing to the
quotient, but we must to  avoids some simple error. as the
deference between bilinear map and bilinear form. and for a
bilinear map there  are two type of kernel that is left and right,
and this two type are equal in the case of symmetric bilinear map.
so in your context   we speak on the left kernel of $f$ as you
have wrote: $kerf=\{x\in V \mid f(x,y)=0\forall y\in V\}$ and this
is a well subspace of $V$ so $\widetilde{V}=V/kerf$ is a quotient
space, but what is $\widetilde{f}$? that is the problem of lifting
$f$ via the space  $\widetilde{V}=V/kerf
\times\widetilde{V}=V/kerf$, this  is the problem :  when
$\widetilde{f} (\widetilde{x},\widetilde{y})=f(x,y)$ has a sense
for all $x,y\in V$? if you don't suppose $f$ symmetric there no
warrant $\widetilde{f}$ to be well defined.
so we answered the question in the case all good, that is $f$
symmetric and so $\widetilde{f}$ is well defined in this case.
in this case we tack $\{e_1,...,e_m\}$ a basis of $kerf$ and we
complete this basis with $\{e_{m+1},...,e_{m+n}\}$  for obtaining
a basis $\{e_1,...,e_m, e_{m+1},...,e_{m+n}\}$ of $V$ then
$\{\widetilde{e}_{m+1},...,\widetilde{e}_{m+n}\}$ is a basis of
the quotient space $V/kerf$.
we prove that $\widetilde{f}$ is not
degenerate bilinear map. let $x\in V$ such that
$\widetilde{f}(\widetilde{x},
\widetilde{e}_j)=f(x,e_j)=0,\;\forall j\in\{m+1,...,m+n\}$, this
applique that $f(x,e_i)=0,\;\forall i\in\{1,...,m+n\}$ (because
for $i\in\{1,...,m\}$  we have $e_i\in kerf$), so $x\in kerf$ and
$\widetilde{x}=0$ this prove that $\widetilde{f}$ not degenerate.
