However, when we use kruskal coordinates, the coordinate system covers the inside of the spherical object (usually taken to be a black hole).
The interior (by which I mean, $r<r_S$) of a Schwarzschild black hole is also a vacuum, insofar as the stress-energy tensor vanishes there.
But it’s a black hole..doesn’t it have mass/energy?
The Schwarzschild metric is a vacuum solution to the Einstein equations. It describes the metric in a vacuum ($T_{\mu\nu}=0$) under the assumptions that the metric is spherically symmetric and stationary. If you model a star as a spherical fluid with a sharp boundary with pure vacuum outside it, then the exterior region would be described by the Schwarzschild metric while the interior of the star (where $T_{\mu\nu}\neq 0$) would be described by the interior Schwarzschild metric. In this case, the parameter $M$ which appears in the exterior Schwarzschild solution is the mass of the star.
One could ask, however, what happens when you assume that $T_{\mu\nu}=0$ everywhere. The most general spherically symmetric, stationary vacuum solution to Einstein's equations is
$$\mathrm ds^2 = -\left(1-\frac{\alpha}{r}\right)c^2\mathrm dt^2 + \left(1-\frac{\alpha}{r}\right)^{-1} \mathrm dr^2 + r^2 \mathrm d\Omega^2$$
where $\alpha$ is an unknown constant. If $\alpha = 0$, then this is just the standard Minkowski spacetime from special relativity. However, it's entirely possible that $\alpha \neq 0$, provided that we remove the set of points with $r=0$ from the spacetime manifold (otherwise, the metric would be undefined there).
One might object immediately by noting that for $r=\alpha$, the metric component $g_{rr}$ is undefined. However, this is what is called a coordinate singularity; it's not actually a problem with the spacetime, just with this choice of coordinates. One can switch to the Kruskal coordinate system to demonstrate this fact. In contrast, the singularity at $r=0$ is a genuine one, and so that set of points must be removed.
So if $\alpha \neq 0$, then what is it? Comparison with the exterior Schwarzschild solution outside of a star provides a physical interpretation - this is what you'd expect to see if a total mass $M\equiv \alpha c^2/2G$ was crushed down to a single point in space (which, to avoid infinities, must be excised from the spacetime manifold). However, it's important to note that at all points (other than the singularity, which again has been removed), $T_{\mu\nu}=0$. This is true both for $r>\alpha$ and $0<r<\alpha$.
[...] from observations black holes are not a single point. Supermassive black holes have a radius of ~15 million miles.
The 15 million miles you refer to is the event horizon of the black hole - the $\alpha$ which appears in my explanation above. Again, for a Schwarzschild black hole, $T_{\mu\nu}=0$ both outside the event horizon and inside the event horizon.
[...] For these black holes there is surely mass and energy, so it seems like we cannot speak of geodesics “inside” the black hole if we are still sticking with schwartzchild space time
One of the reasons I keep using the qualifier Schwarzschild to refer to a black hole is because this metric is the simplest possible model for understanding the phenomenon. It is not necessarily a realistic model for an actual black hole, which would generically be e.g. rotating and surrounded by an accretion disk which is slowly falling below the event horizon. Rather, you should think of it as a toy model for understanding the nuances of the event horizon and other highly non-intuitive phenomena. Adding a steady influx of mass or some other more complex features would only obscure these issues.
