I'll first recapitulate the why calculus works for Riemannian manifolds, and then present why I believe the same does not work for Pseudo-Riemannian manifolds. I'd like to know where my current model of the situation breaks down.
A differentiable manifold $M$ of dimension $n$ is a topological space $M$ equipped with an open cover $A$ (the atlas) such that for each chart $C_i \in A$, we have a homeomorphism $\phi_i: C_i \rightarrow U_i\subseteq \mathbb R^n$ [we take the subspace topology on $C_i$]. Now, for the manifold to be a differentiable manifold, we also ask that for all charts $C_i, C_j$ such that $C_i \cap C_j \neq \emptyset$, the map $\phi_j \circ \phi_i^{-1}: U_i \subseteq \mathbb R^n \rightarrow U_j \subseteq \mathbb R^n$ is smooth: That is, our change-of-coordinates maps are smooth maps.
At this point, we can define calculus on the manifold. We now define the tangent space at a point $p \in M$, where $p \in C_p$ [$C_p$ is in the cover]. We have $\phi_p: C_p \rightarrow U_p \subseteq \mathbb R^n$ as the local chart. The tangent space $T_p M$ is the collection of all smooth curves $\{ c: [-1, 1] \rightarrow C_i \mid c(0) = p \}$. But we want to impose an equivalence relation on these curves, to consider curves that have equal velocities at $p$ to be equivalent curves. So we define $c_1 \sim c_2 \iff (\phi_p \circ c_1)'(0) = (\phi_p \circ c_2)'(0)$ . One can check that this gives a vector space of dimension $n$.
- Note that we need the differentiable structure of $\mathbb R^n$ for this definition to work. In particular, we need ideals such as limits behave 'the same' in both $C_p$ and $U_p$ [the coordinate system for $C_p$]. We are guaranteed this will happen because $C_p$ is homeomorphic.
- We know that all of these calculations are not dependent on the coordinate system because we have assumed that changing coordinates is smooth. Hence we always switch to an alternative coordinate system and suffer no change. The Jacobian of the change-of-coordinates map $\phi_j \circ \phi_i^{-1}$ induces an isomorphism of tangent spaces.
When we now go to the Riemannian case and impose an inner product $\langle \cdot \vert \cdot \rangle_p: T_p M \times T_pM \rightarrow \mathbb R$ at each point $p$, we are guaranteed that this inner product does not disturb the topological considerations needed by the above differentiable structure thanks to equivalence of norm in a finite dimensional vector space. The topology induced by $\langle \cdot \vert \cdot \rangle_p$ is the same as the standard topology, and hence our manipulations of calculus 'are compatible' with our choice of inner product.
If we now consider the Pseudo-Riemannian case, the bilinear form $\{\cdot \mid \cdot\}: T_p M \times T_p M \rightarrow \mathbb R$ need not be positive definitite (and hence need not induce an inner product). The way this bilinear form 'sees the space in local coordinates' can be pretty weird.
For example, let us take Minkowski space $M \equiv (\mathbb R^4, \{ \cdot \mid \cdot \})$ such that $\{ \mathbf p, \mathbf q \} \equiv -p_0 q_0 + p_1 q_1 + p_2 q_2 + p_3 q_3$. Then for vectors $\mathbf v = (t, x, 0, 0), \mathbf w = (x, t, 0, 0)$, we have $\{v, w \} = 0$, while the regular inner product $v \cdot w = 2xt$. So this bilinear form does not induce any reasonable norm (and hence no topology).
Given all of this, why is it legal for us to still perform computations on the tangent space in a Pseudo Riemannian manifold? As far as I can tell, the 'bilinear form' structure is in no way compatible with the differentiable structure of $\mathbb R^n$.