In some discussions of metric tensors $g : TM_p \times TM_p \rightarrow \mathbb R$ (and even of pseudo-metric tensors, for instance in this answer), and of the corresponding first fundamental forms ${\rm d}s^2$, these (mathematical) objects are referred to as "metric".
Is this serious and correct terminology?
Is it really true that metric tensors, or (even) pseudo-metric tensors, and/or the corresponding first fundamental forms are metrics (in the specific well-known sense), by which accordingly $g : X \times X \rightarrow \mathbb R$ for some suitable set $X$, such that $\forall x, y, z \in X$:
$g[~x, y~] \neq 0 ~ ~\Leftrightarrow ~ ~ x \not\equiv y ~$ (distinguishability of distinct elements),
$g[~x, z~] \le g[~x, y~] + g[~y, z~]$ (subadditivity; triangle inequality)
?
Yes, it is true, but note that the metric tensor $g$ defines the distance function on $TM$ and hence a norm, it is not itself the distance function.
In particular, for $TM$ we can define
$$d(\vec{x}, \vec{y}) \equiv \sqrt{g(\vec{x} - \vec{y}, \vec{x} - \vec{y})} \quad\forall \vec{x}, \vec{y} \in TM$$
It's relatively easy to show that if $g$ is a metric (as opposed to a pseudometric) this definition of $d$ works as you would expect.
Note that I am answering about the relationship between $g$ and $d$ on $TM$: $g$ also makes $M$ a metric space, but that's a slightly different argumentt.