A bilinear form $B: V \times V \to \mathbb R$ is nondegenerate if the map $V \to V^*$ defined by $v \mapsto B(v,-)$ is injective for all $v \in V$.
Take $V= \mathbb R^2$ and take $B$ to be the standard inner product on $\mathbb R^2$.
Take $v=(1,0)$. Then $B((1,0), (0,1))=1 \cdot 0 +0 \cdot 1 =0$. But then the map $B((1,0),-)$ is not injective since $(0,1) \mapsto 0$ but $(0,1) \ne (0,0)$.
What is wrong here?