It might help to think of an example. A simple is example is dust, specifically a collection of particles of a fixed rest mass, all at rest with respect to each other, and we can consider uniform dust, so they are equally spaced.
If that's the only thing in our universe, then there is no momentum or stress in the frame of the dust and the energy is just the rest-energy, so the mass density and the energy density are simply proportional.
So that's our example. Now let's look at the stress energy tensor. We have a $T^{\mu\nu}$ as the flux of four-momentum $p^{\mu}$ across a surface of constant $x^{\nu}$. A surface of constant $x^0=ct$ is a surface of constant $t$. A flux is a per-area thing. So you can imagine a bit of area/volume in the $t=const$ plane/hyperplane, say a rectangle/box with size $\Delta x \Delta y \Delta z$, if the box is bigger you get more flux. We can draw the worldlines of the particles and count how many pierce through this piece of the $t=const$ hypersurface, and once the piece is small enough, the result is proportional to the size of the volume. So that propotionality constant is the particle density. If we multiply that by the mass per particle, we are now counting mass that pierces that portion of the $t=const$ surface, and the proportionality constant is mass density. If we multiply that by the $c^2$, we are now counting energy that pierces that portion of the $t=const$ surface, and the proportionality constant is energy density.
It's a number that tells us how much the $p^0$ component of pierced a piece of a surface of constant $x^{\nu}$ divided by the size of the piece. Why is it called a flux?
Fluxes are (thing/area)/time. If you set up a surface of $x^1=const$ then you can make a piece of that surface with a $\Delta t$, and a $\Delta y$ and a $\Delta z$ and see how much of of your thing hits the piece of the $x^1=const$ surface inside your patch, and it is obviously propoertional to the duration of the patch (has to hit at within the time interval) and the area of the patch. So the rate per area is a measure of the constant of proportionality between how many pierced the piece of the $x^1=const$ surface and the "volume" $\Delta t \Delta y \Delta z$ of the piece.
So flux is the version when you have a "volume" $\Delta t \Delta y \Delta z$ for a piece, and density (stuff per volume $\Delta x \Delta y \Delta z$) is the name when you have a piece of a $t=const$ surface. Rightly they are the same exact concept. So we either call them both densities or call them both fluxes or we call them density-flux or flux-density.
It is called a flux because it's the thing you multiply by the "volume" $\Delta t \Delta y \Delta z$ to get how much of the thing pierced your piece of the hypersurface $x^1=const$. When the "volume" is an actual volume $\Delta x \Delta y \Delta z$ then historically we called that constant a density before we knew that spacetime is legitimate.
Recognizing that flux is a rate per area, and that this generalizes to density for $t=cosnt$ hypersurfaces is all you need to understand it. Now you can do particle flux, mass flux, energy flux, etc.
edit
So that's what flux and density are, and they are the same concept. Let's address your specific questions one by one:
$T^{00}$ should be the flux of energy through space, right?
No, $T^{00}$ is the flux of energy through a surface of $t=const$, or more rightly $$\int\int\int_{t=const}T^{00}dx^1dx^2dx^3,$$ should give you how much energy passed through your region, so locally $T^{00}$ is a constant of proportionality that scales $\Delta x \Delta y \Delta z$ up to a little bit of energy. Historically we'd call it energy density, but in relativistic physics we call it a flux to acknowledge that there is nothing different in principle between that constant and the constant you multiply by $\Delta t \Delta y \Delta z$ to see how much stuff flows through an area $\Delta y \Delta z$ in a time interval $\Delta t$.
I think I'm not understanding is the usage of the word flux here.
A flux is a constant of proportionality you multiply by $\Delta t \Delta y \Delta z$ to see how much stuff flows through an area $\Delta y \Delta z$ in a time interval $\Delta t$, all in a surface of $x=const$. So to be fair to all directions of spacetime, you can pick any surface like $dx^{\nu}=const$ then take $c\Delta t \Delta x \Delta y \Delta z/ \Delta x^{\nu}$ to quantify how much of that infinite surface $dx^{\nu}=const$ you have and the constant of proportionality that you multiply $c\Delta t \Delta x \Delta y \Delta z/ \Delta x^{\nu}$ by is called the flux.
However, what all the resources I've looked at say is that the flux of energy through space would be the energy density, which isn't at all the equation above.
An integral with $dxdydz$ looks exactly like you are integrating a density, and a density on a $t=const$ surface is exactly a flux for $dx^{\nu}=const$ surface for the special case where $\nu=0$.
If $T^{00}$ were to be the energy density $\rho$, wouldn't it make more sense to change the definition of the stress energy tensor, such that:
The four-momentum $p^{\mu}$ is the flux of $T^{\mu\nu}$ through a surface of constant $dx^{\nu}$.
OK. Sometimes people call the flux the rate, the thing per area per time, or the thing per volume. But sometimes they call the thing integrated the flux. This is unfortunate. Confusing the two is like confusing an energy and an energy density. The stress-energy tensor is telling you the rate, the per volume quantity:
$$p^{\mu}=\int\int\int_{t=const} T^{\mu 0} dx dy dz.$$
Instead of choosing a $t=const$ surface and seeing how much $p^\mu$ crossed it, you could pick an $x=const$ surface and get:
$$p^{\mu}=\int\int\int_{x=const} T^{\mu 1} cdt dy dz.$$
Or you could pick a $y=const$ surface and get:
$$p^{\mu}=\int\int\int_{y=const} T^{\mu 2} cdt dx dz.$$
Or you could pick a $z=const$ surface and get:
$$p^{\mu}=\int\int\int_{z=const} T^{\mu 3} cdt dx dy.$$
In each case, the integral is telling you how much $p^\mu$ crosses your hypersurface. And it turns out this is enough to handle any hypersurface, in particular if you pick any surface that locally looks flat, you can combine these rates per area (or densities) $T^{\mu 0}$, $T^{\mu 1}$, $T^{\mu 2}$, and $T^{\mu 3}$ to find out how much $p^\mu$ flows across your arbitrary surface.
