Comparing two spacetimes Consider two spacetimes with the same manifold $M$ but distinct metric $g,g'$. How do I identify a point $p \in M$ in two spacetimes? Specifically, if I give the coordinates of the point in one spacetime, can I find it in the other spacetime? 
E.g.  Let $M=\mathbf{R}^4$, If a point $p\in \mathbf{R}^4$ in Minkowski spacetime $(\mathbf{R}^4,\eta)$ is given in polar coordinates as $(t_0,r_0,\theta_0,\phi_0)$, does it mean in Schwarzschild spacetime with Schwarzschild coordinates, it is the point with coordinates $(t_0,r_0,\theta_0,\phi_0)$ as well? Do they use the same coordidnate chart?
I think maybe the coordinate charts have to agree in order to do that, but I am not sure whether there is a coordinate-independent way of doing that. If not, how do we check whether the two coordinate charts agree in general?
 A: If you have two different spacetimes, there is no standard way to identify their points. In particular, the Schwarzschild spacetime and Minkowski space aren't even homeomorphic (one has the topology $\mathbb R^2 \times S^2$ while the other is $\mathbb R^4$), so at best you'd have to define it on a subset of both spacetimes. 
The most common way to identify points on two spacetimes is to consider a diffeomorphism between the two, or a local diffeomorphism if they are not themselves diffeomorphic. Consider a neighbourgood $U_1 \subset M_1$ and $U_2 \subset M_2$, such that $U_1 \cong U_2$ (this is always possible since you can just pick a ball), then you can just define a diffeomorphism $\phi : U_1 \to U_2$ to map coordinates from one manifold to the other. In this case, it would be, for a coordinate chart $\phi_1$ and $\phi_2$
$$\phi_2 \circ \phi \circ \phi_1^{-1}$$
which maps coordinates from $U_1$ to $U_2$.
You may also want something a little stronger, such as an isometry, where the metric is preserved by the transformation, written as $g_1 = \phi^* g_2$. In this case we say that the two spacetimes are the same if they are isometric, since the metric tensor is the only quantity we really care about for spacetimes. The same way you can define a local isometry rather than a full on isometry here.
One area where such things are useful is to consider the possibilities a spacetime may evolve to. Consider two spacetimes which are isometric up to a certain point in time, so that there's isometric subsets $U_1$ and $U_2$. Then it makes some sense to compare the coordinates of these two spacetimes in the region where they are isometric. It may also be useful to consider the possibilities of what the spacetime may be, since we are likely to know the metric of a spacetime only in a small region and there are different spacetimes that may admit the same local metric.
On the other hand, if you don't really have isometric spacetimes, while you can compare coordinates, these choices will be somewhat arbitrary. "Nice" manifolds (which spacetimes are) are homogeneous : there's always a diffeomorphism taking a point of $U_1$ to any other point of $U_2$.
A: The metric doesn't really play a role in your question. The reason why it appears I guess is because you often consider just a single coordinatization of your space (if possible, like on $\mathbb R^4$), and you choose one in which the quantities that you are interested in, like the metric, are easy to describe.
I guess that what you really want is given two charts, how to see if a coordinate tuple in one chart describes the same point as a coordinate tuple in the other.
The general answer may be a bit disappointing: let's go to the definition of a chart on a manifold. A manifold ultimately is a topological space with extra structure that is defined in terms of homeomorphisms between open subsets and open subsets of some $\mathbb R^n$ (the open subsets of the manifold are charts, the homeomorphism provide local coordinates), satisfying some properties. 
Since that's essentially the full definition, you cannot say more than that two coordinate tuples belonging to different charts describe the same point if the inverse homeomorphisms map them to the same point.
