As an American president once said "it all depends on what the meaning of the word is is". If you look at the definition, it is clear that Riemannian metrics, as defined, are not distance functions. When you say that a (connected) Riemannian manifold $(M,g)$ is a metric space, you are actually saying that there exists a certain functor from the category of Riemannian manifolds to the category of metric spaces:
$$
\Phi: (M,g)\mapsto (M, d_g)
$$
where $d_g$ is the Riemannian distance function. (If you do not know what functors and categories are, just think of functors as maps and categories as sets, even though this is, strictly speaking, false.) This functor is like a conversion procedure from, say, kilograms to pounds (except, it is not invertible). The functor $\Phi$ is defined by first assigning a quantity, which we call length $L_g(p)$ to each (rectifiable) path $p$ in $M$ and then minimizing.
The step $g\mapsto L_g$ makes sense even for pseudo-Riemannian manifolds, just you get some paths of negative length. It is the 2nd step which fails: If you try to minimize, you will (frequently) get $-\infty$, which is not particularly useful.
Remark. What I said above about isometries: Both categories of Riemannian manifolds and metric spaces have their own notions of isometric maps and these two notions do not correspond to each other under the functor $\Phi$. For instance, a Riemannian geometer would think of the circle of length $2\pi$ isometrically embedded into $R^2$ (with the image of the embedding being the standard unit circle). But this embedding does not preserve distances between points! This is another reason I do not like to use the word is for the relation between Riemannian manifolds and metric spaces. However, assuming that you know what functors are, otherwise, just ignore this: $\Phi$ does become a functor if we use 1-Lipschitz maps between spaces as morphisms for both Riemannian manifolds and metric spaces.
Now, to your question of why do we call pseudo-Riemannian metrics metrics, it is all matter of habit and tradition. You can think of three different worlds:
Metric geometry
Riemannian geometry.
Pseudo-Riemannian geometry.
They all have their notions of metrics (and isometries), but these notions have different meanings. It is as if people who speak different languages can occasionally use the same word, but it has different meaning in these languages. My favorite example is the word application, which has different meaning in English and in French.
It fails even at the pointwise level: in the Riemannian case,
$$\tag{1} \sum_\beta \tau_\beta g_\beta$$
is always positive definite since $g_\beta >0$, $\tau _\beta \ge 0$ and $\tau_{\beta_0} >0$ for some $\beta_0$. But if each $g_\beta$ is only non-degenerate, then (1) might give you a degenerate symmetric two tensor.
Related question about the existence of Puesdo Riemannian metric: here
Best Answer
A necessary and sufficient condition for a smooth $n$-manifold $X$ to admit a metric of signature $(r, s)$ is that the tangent bundle admits a direct sum decomposition into a bundle of rank $r$ and a bundle of rank $s$, or equivalently that the tangent bundle admits a reduction of the structure group to $GL_r \times GL_s$.
When either $r$ or $s$ is equal to $1$ this is equivalent to the existence of a nonvanishing vector field, which is automatic when $X$ is noncompact and governed by whether $\chi(X) = 0$ otherwise, by the Poincaré–Hopf theorem.
In general I think it's a difficult question to determine when this reduction is possible. There are necessary conditions coming from characteristic classes which show that, for example, an even-dimensional sphere $S^{2n}$ does not admit a metric of any indefinite signature. A sufficient condition is that the manifold smoothly fibers over an $r$-manifold (or an $s$-manifold).