Recall that a coordinate system is for $M$ is given by an open set $\Omega\in \mathbb{R}^n$ and a map $\psi: \Omega\to M$ that is a diffeomorphism between $\Omega$ and its image. Then the inverse $\psi^{-1}$ gives a local coordinate system $\psi^{-1}: \psi(\Omega) \to \mathbb{R}^4$.

In the Schwarzschild case it is useful to think in terms of the corresponding map $\psi$ and not the actual coordinate functions $\psi^{-1} = (r,t,\theta,\psi)$. In other words, what you want to do is look at the map that sends points in $\mathbb{R}^4$ with the coordinate value $(r,t,\theta,\psi)$ to points on the Schwarzschild manifold.

Looking at it this way, the first problem with the Schwarzschild coordinate is that when $r = 2M$,
$$ \psi(2M,t_1,\theta,\psi) = \psi(2M,t_2,\theta,\psi) $$
for any pair of $t_1, t_2$. This automatically implies that $\partial_t\psi =0 $ and hence $\psi$ is not a diffeomorphism.

Another way to say this is that when $r = 2M$, the mapping $\psi$ corresponding to the inverse of the coordinate functions is independent of $t$; for every $t$, the image of $\psi(2M,t,\cdot,\cdot)$ is a fixed 2-sphere in the Schwarzschild manifold (the bifurcate sphere for the past and future event horizons).

This is analogous to the situation of the function
$$ u: \mathbb{R}^2 \to \mathbb{S}^2\subset \mathbb{R}^3 $$
where
$$ (\theta,\phi) \mapsto (\cos\theta, \sin\theta \cos\phi, \sin\theta\sin\phi)$$
This is a smooth map and is (up to some notational changes) the common way of parametrizing the sphere (with $\theta\in [0,\pi]$ and $\phi\in [0,2\pi)$). When $\theta = 0$ or $\pi$, you get the north/south pole *independent* of $\phi$.

Wait, but you ask: the metric for the sphere in the $(\theta,\phi)$ coordinates degenerates, but it doesn't blow up! Why is it that the metric in the Schwarzschild case blows up?

The answer is that the problem with the $t$ coordinate is *not* the only problem with the coordinate system. The $r$ function can be geometrically defined as the *area-radius* function on the spherically symmetric manifold that is the Schwarzschild solution. It turns out that the function $r$ *has a critical point exactly on the bifurcate sphere*. The critical point is a "saddle point". Away from the bifurcate sphere $r$ is non-critical, and so we can define the smooth function $\psi$. But the criticality of $r$ (which is part of the inverse of $\psi$) implies that $\psi$ cannot be extended to a differentiable map near where the image is the bifurcate sphere.

Putting everything together: there are two things that go wrong with the Schwarzschild coordinates at $r = 2M$.

- From the point of view of the mapping $\psi:\mathbb{R}^4\supset\Omega \to M_S$: when $r = 2M$ the function becomes constant in $t$, and hence is not a diffeomorphism.
- Furthermore, since the "inverse" function $r:M_S\to\mathbb{R}$ has a critical point on the bifurcate sphere [which as we recall corresponds to exactly when $r = 2M$ in the Schwarschild coordinates], this means $\psi$ is also not differentiable at $r = 2M$, and so is not a diffeomorphism.

The first gives rise to the fact that one of the metric coefficients in the Schwarzschild metric being 0, the second gives rise to the fact that one of the metric coefficients become infinite.