Why is $\langle \operatorname{grad} f, X\rangle_g$ independent of the metric on a Riemannian manifold?

Yes, the reason is simple. By definition, $\operatorname{grad}f$ is the vector field (metric-dependent) so that $\langle\operatorname{grad}f,X\rangle_g = df(X)$, and the right-hand side is independent of the metric. But the vector field itself does depend on the metric. :)

Absolutely. The exterior differential of a function, $df$, is a one-form which is defined independently of metric: $df(X) = X(f)$. In coordinates, $df = \partial_if dx^i$.

Choice of a nondegenerate metric induces an isomorphism between vector fields and $1$-forms by pairing: $g(X)(V) = \langle X,V\rangle$. In coordinates, this pairing corresponds to multiplying by the inverse of the metric, or "raising indices." This isomorphism is also known as a musical isomorphism: if $X$ is a vector field, then the one-form $X^\sharp$ is defined by $X^\sharp(V) = \langle X,V\rangle$.

The gradient $\operatorname{grad}(f)$ is defined to be the metric dual of the one-form $df$, i.e., $\operatorname{grad}(f) = (df)^\sharp$, so if you pair $\operatorname{grad}(f)$ with an arbitrary vector field $X$, you get $df(X) = X(f)$, which does not depend at all on the metric. The gradient is constructed so that, in pairing, all mentions of the metric cancel.