Addition of gravitational fields in general relativity According to Newtonian gravity, when dealing with multiple sources of gravity, the net gravitational field equals the sum of the individual gravitational fields of each source.
Does the same hold for general relativity (GR)?
Because gravity is described by a spacetime metric, and not by a field, is it accurate to say that given multiple sources of gravity, each source corresponding to some metric, the "net metric" at a given point equals the sum of the individual metrics of each source?
 A: While I am no expert on GR - here's another perspective.
While the forces in Newtonian gravity do follow super-position, the potential energy does not: the force experienced by a test mass in the presence of multiple sources is same as the sum of forces experienced when only individual sources were present but the potential energy in the presence of multiple sources isn't the sum of the potential energies of the individual cases - there is an additional contribution - the mutual potential energy of the multiple sources.
Note that this "energy of the sources"$^1$  is a constant contribution regardless of where the test mass is. Further, the field depends on the gradient of the potential so constant contributions don't matter and so fields still add up linearly.
Now, in GR, energy acts as a source of gravitational force$^2$. From this alone, it follows that when we try and super-position the solution of individual sources, a new gravitational source is activated - the mutual energy of the sources. This means the metric being calculated for the source as a whole will be different$^3$ than the "sum of the metrics"$^4$ of the individual sources.
In one-line, this lack of superposition is a synonym for non-linearity, as tersely put by @ConnorBehan.

$^1$ aka. binding energy
$^2$ it contributes to the stress-energy tensor, which is the source term in the Einstein field equations.
$^3$ What about the gradient logic of Newtonian gravity? Why doesn't that apply here? Well, in GR the quantity of interest, the metric, isn't a simple gradient or any linear function of the sources so it doesn't apply.
$^4$ I am not sure what that would mean but sum in some sense
A: Nope. GR is a non-linear theory.
A: One of the most interesting, and complicated, features of general relativity is the fact it is a non-linear theory, i.e., adding solutions together won't yield a solution.
One example of that behavior is a Schwarzschild black hole, which means a black hole with no charge and no rotation (a spinning black hole is a bit more complicated, but it would work as well). The Schwarzschild solution is what is known as a vacuum solution: there is no matter in the spacetime. At any point in spacetime you look, there won't be matter. Still, there certainly is gravity.
One of the pictorial ways of interpreting this is by noticing that the gravitational field itself possesses "energy" (in quotation marks, because the notion of energy in general relativity is complicated, as mentioned in this blog post by Sean Carroll). By means of $E=mc^2$, having energy means, in some sense, having mass, and hence the gravitational field is a source of even more gravitational field. The gravitational energy "creates" more gravity, which leads to more gravitational energy, and then more gravity and... And this is essentially what we call non-linearity. The effects start piling up on each other and the description gets quite complicated. Far more complicated than what one has in Newtonian gravity or Maxwellian electromagnetism, both of which are linear theories.
Notice then that if you add two solutions together, you'll be increasing the amount of gravitational energy. That is a source of more gravity, and hence you'll need to account for this extra gravity, which was not present in the two original solutions.
Disclaimer: notice that this "energy creates more gravity" view is pictorial, and meant only to bring more intuition. There is no way of assigning an "adequate" notion of energy to the gravitational field (see the Sean Carroll blog post for some more detail). While this picture can be used as a way of getting intuition and interpretation, it does have limitations and should be taken with a grain of salt.
Black Holes are made of Vacuum
I noticed this bit of the answer caused some discomfort on the comments, so maybe I should add some more resources on it. I believe Kip Thorne is someone particularly famous who quite often mentions how black holes are made of warped spacetime, instead of compact matter. His comments appear in this site now and again. Here are some instances:

*

*Why does Kip Thorne claim spacetime warping itself contains energy?

*Mass without Matter?

*If a black hole is just warped spacetime, then where is the electric charge?
I should also have added before that all of my answer should be understood in the context of General Relativity, which means I'm neglecting all quantum effects. Within the framework of General Relativity, spacetime is a differentiable Lorentzian manifold, which means it must have a well-behaved metric at all points. This prevents the singularity of Schwarzschild spacetime from being a point in the manifold, since a curvature scalar blows up "there". Hence, in the description provided by General Relativity, there is not a single point in Schwarzschild spacetime where there is matter. All points are at vacuum.
"But near the singularity, quantum gravity effects should kick in and—" I agree. This description is not necessarily final, and most likely it will be modified by quantum effects. However, it is important to distinguish what happens in the actual Universe—in which the Schwarzschild solution doesn't even exist, since we had a Big Bang and we have a positive cosmological constant—and what is described by General Relativity. I discussed similar issues (the difference between theory and reality) in this post about Classical Electrodynamics. It is one thing to ask whether an actual black hole corresponds to complete vacuum, and another thing to ask whether a GR solution is a vacuum solution.
A: GR is a non-linear theory, and that is enough to answer the OP's question. Regarding how a test particle possessing mass affects the gravitational field in GR, search the literature on "Back Reaction".
A: According to Newtonian gravity, it doesn't account for the strong and weak equivalence principle, where you cannot just aggregate all the sum of their gravitational fields and derive a solution, it doesn't make sense nor does it have any pragmatic solution. The same as Einstein's spacetime metric version of gravity. To be clear about understanding the significance of "gravitation" there is no one-standard universal law unless you incorporate the SEP, and WEP. The Newtonian version of gravity coupled with Herr Einstein's field equations can best be described, as the force of attraction between two masses divided by the square of their distance in spacetime curvature. The second universal axiom, the SEP and WEP are directly proportional to the product of the sum of their masses or matter fields, and inversely proportional to their distances. The third axiom, gravity is a nonlinearized scalar-tensor-vector matter field.
