Known methods to detect coordinate singularities? MOTIVATION
In Riemannian geometry, when one writes the metric tensor in a particular coordinate system, certain 'fake' singularities might appear that have little to do with the geometry near them.
For instance, on the sphere $S^2$, one has (with appropriate labelling)
$$ds^2 = d\phi^2 + \sin^2(\phi)d\theta^2$$
and so it looks as though the spherical metric is singular at $\phi = 0$ or $\phi = \pi/2$, but we know it is actually not. However, the only reason we know this is because we have other descriptions of this metric that are evidently not singular anywhere.
SETUP
Suppose you have a smooth manifold (let's say three-dimensional) $M$ and you are given a 
local description of a metric
$$ds^2 = g_{ij}dx^idx^j$$
on a coordinate patch $U \subseteq M$. If you want, you can assume this $U$ is dense in $M$, like in the sphere. Suppose that the metric looks singular near the points in $M\backslash{}U$.
Here, I mean the same type of singularity as in the sphere, that is, some crucial terms $g_{ij}$ vanish when you go away from $U$, so that the matrix $[g_{ij}]$ would have zero determinant.
Question: What are the known methods for not just detecting, but actually proving that the metric is not singular and therefore can be extended to the whole $M$?
 A: I think there are two slightly different questions here. The first question is the one you specifically asked: If we are given the manifold $M$ and a metric defined on a dense coordinate chart $U\subseteq M$, what is a sufficient condition for the metric to have a smooth extension to all of $M$?  This is basically pretty easy to answer: just find, for each point in $p\in M\smallsetminus U$, a smooth chart on a neighborhood $V$ of $p$, and transform the metric into the new coordinates on $V\cap U$.  If the transformed metric extends smoothly to $V$, then $p$ was just a coordinate singularity. If $M$ is compact, you can do this with finitely many charts, and often finitely many will do in the noncompact case as well.
The more subtle question arises when $M$ is not known in advance. In this case, we have a smooth metric on an open manifold $U$, and we want to know whether there is some smooth manifold $M$ properly containing $U$ and a smooth metric on $M$ that extends the given metric on $U$.  (When this is the case, the given metric on $U$ is said to be extendible.) This is a much more difficult question to answer.  There are some obvious necessary conditions for a metric to be extendible: for example, the metric must be geodesically incomplete, and all of the sectional curvatures have to remain bounded as you approach the boundary of $U$.  But I would be surprised if there are any simple sufficient conditions. 
To see how strange things can get, let $U$ be the universal cover of $S^2$ with the north and south poles removed, with the metric obtained by pulling back the standard round metric via the projection $U\to S^2$.  (You can visualize this as an "infinite-sheeted orange peel.") Then $U$ is diffeomorphic $\mathbb R^2$, and the pullback metric is incomplete and has constant sectional curvature, but it's not extendible.
A: Qiaochu's comment misses the point: Some metrics have true singularities that cannot be removed by a change of coordinates.  
I don't know about general metric spaces, but in good-old 4-D space-time,  a  sufficient condition for a true singularity to exist at point $P$ is that the limit as one approaches the singularity of the trace of the Ricci tensor goes to infinity  as one approaches $P$. 
this is a fairly powerful test, but I don't remember if this is also a necessary condition; I think it is given sufficient differentiability of the metric.
It turns out that it is not sufficient (although for the Schwartzchild solution this scalar is $\frac{48G^2 M^2}{c^4r^6}$ so it does, contrary to one of the comments, tell you that the singularity at $r=0$ is true.) For metrics that are at least semi-Reimannian, the necessary and sufficient condition for non-singularity is that in the Ricci Decomposition of the curvature, all three of the following are non-singular:  The scalar curvature (my original criterion); the semi-traceless part of the curvature tensor, and the fully traceless Weyl tensor. This applies in any number of dimensions above 2. I think the charged Kerr-Newman solution might pass the scalar curvature condition but fail the semi-traceless part condition at the ring of singularities.
