Let there be $f:V\times V \rightarrow F$ over a finite vector space, which for a certain base $(w) = \{ w_1, w_2, ... , w_n \}$ has a nonsingular representing matrix (e.g. $[f]_w$ is nonsingular).
How can I show that for every linear transformation $l:V\rightarrow F$ there exists a vector $v_0 \in V$ so that $l(u) = f(u, v_0)$ for each $u\in V$?
Consider the map $V \to \text{Hom}_F(V,F), v \mapsto (u \mapsto f(u,v))$ and show that your assumption on $f$ implies that this map is an isomorphism. In fact, it suffices to show that this map is injective, as $V$ is assumed to be finite dimensional (and $\text{Hom}_F(V,F)$ has the same dimension as $V$).
Indeed, if $v \in V$ is mapped to $0 \in \text{Hom}(V,F)$, this means $f(u,v) = 0$ for all $u \in V$. By your assumption, this is only possible if $v = 0$, hence the above map is injective and due to dimension, it is surjective and this is what you want.