In O'Neill's Semi-Riemannian Geometry book, he calls a symmetric bilinear form over a vector space $V$ nondegenerate if $\forall v \in V, v \ne 0 \implies \exists w \in V, \langle v,w \rangle \ne 0$. He then defines the following:
A metric tensor $g$ on a smooth manifold $M$ is a symmetric nondegenerate $(0,2)$ tensor field on $M$ of constant index.
Here he's calling a tensor field nondegenerate. A $(0,2)$ tensor field can be thought of as a bilinear form on the module of vector fields over smooth functions, thus we can get a notion of nondegeneracy from just taking the original definition and generalizing to modules. In particular, we would say a symmetric $(0,2)$ tensor field $T$ is nondegenerate if (denoting the ring of smooth functions on a manifold $M$ by $F(M)$): $$\forall f \in F(M), f \ne 0 \implies \exists g \in F(M), T(f,g) \ne 0.$$
Unfortunately this doesn't induce a nondegnerate bilinear form at each $p \in M$.
We can see this by considering the $(0,2)$ tensor field on $\mathbb{R}$ defined pointwise as $g(p) = p^2 dx \otimes dx$. Then the induced bilinear form on $T_0\mathbb R$ is clearly degenerate, but $g$ is nondegenerate.
Thus should we say a symmetric $(0,2)$ tensor is nondegenerate according to the above definition, or if it induces a nondegenerate bilinear form at each $p \in M$? For a metric tensor, I believe it should be the latter, though it seems like the former is more correct (uniform in notation).
A bilinear form on a f.d. vector space $B : V \times V \to k$ is nondegenerate iff the induced map
$$V \ni v \mapsto B(v, -) \in V^{\ast}$$
is an isomorphism; it happens that in the f.d. case this is equivalent to requiring that it be injective (which is what you've written) but what we really want is the isomorphism. This definition globalizes correctly: you want a tensor field inducing an isomorphism between the tangent and cotangent bundles (the musical isomorphism), and this is equivalent to requiring pointwise nondegeneracy.