A pseudo-Riemannian “metric” is a nondegenerate quadratic form on a real vector space $R^n$
A pseudometric space $(X,d)$ is a set $X$ together with a non-negative real-valued function $d\colon X \times X \longrightarrow \mathbb{R}_{\geq 0}$ (called a '''pseudometric''')
A bilinear form is simply a linear map $\langle -,-\rangle\colon V \otimes V \to k$ out of a tensor product of k-modules into the ring k (typically taken to be a field).
Minkowski metric is described as a pseudo-Euclidean metric, but not only is it not positive definite, it defines a pseudonorm which takes negative values.
When we talk of a (pseudo-)Riemannian metric, we are not simply talking of the (pseudo-)metric on a vector space. We are talking of a (pseudo-)metric field. The (pseudo-)metric is a property of the tangent space at a point on the manifold; a (pseudo-) metric is strictly a field, that is it is function from the points on the manifold to the tangent (tensor) space at each point.
For metric space we don't need that space should be a real vector space, it can be just a set with a a non-negative real-valued function that respect 3 rules (symmetry, subadditivity/triangle inequality)
nondegenerate quadratic form arise a requirement for a vector space, not exist for a pseudo-metric space
Question: why use a complicated pseudo-Riemannian space solution that include a bilinear form instead of a very more simply, pseudo-metric metric idea ?
Why use a twisted pseudo-Riemannian contortionism into Minkowski metric?