Q: Is a metric diagonalisable if it can be written as $g_{\mu\nu} = \delta_{\mu\nu}$?
A: Yes. In fact, a metric is real and symmetric by definition so it is diagonalisable no matter what.
Q: Is a metric diagonalisable only if it can be written as $g_{\mu\nu} = \delta_{\mu\nu}$?
A: No. The Schwarzchild metric is diagonalisable (as described above) but it has curvature which the delta metric does not.
Q: Is a metric diagonalisable only if it can be written as $g_{\mu\nu} = \delta_{\mu\nu}$ at a point?
A: Still no because metrics for space-time rather than just space are diagonalisable but they contain eigenvalues of both signs. So one can at most hope to write them as $g_{\mu\nu} = \eta_{\mu\nu}$ at a point.
Q: Is a metric diagonalisable only if it can be written as the flat metric appropriate for its signature at a point?
A: No and again Schwarzschild provides a counter-example. At the horizon radius $r = 2m$, one of the eigenvalues is zero meaning there is no change of co-ordinates which can rescale it back to $\pm 1$. Nothing unusual happens here if we look at invariant quantities so this is only an artifact of our description. Co-ordinate singularities like this are common because most manifolds cannot be covered by a single co-ordinate patch. A more drastic singularity occurs at $r = 0$ signalling the need for some new physics.
Q: Is a metric diagonalisable only if it can be written as the flat metric appropriate for its signature at a non-singular point?
A: Yes and this is a theorem. Namely, the existence of normal co-ordinates.
None of these statements make reference to time reversal. To do that, we need the space-time manifold to be orientable. When it is, gravity on its own respects time reversal invariance but other theories that we couple to it (or place on the ambient space-time) might not.
