For any matter/energy distribution we can in principle assemble it from point particles. So the stress-energy tensor of the whole system can be expressed as a sum of the stress-energy tensors of the point particles. The reason this helps is that the stress-energy of a point particle is very simple. It is:
$$ T^{\alpha\beta}({\bf x},t) = \gamma m v^\alpha v^\beta $$
at the position of the particle and zero everywhere else. The variable $v$ is the velocity vector $(c, \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt})$ i.e. it is the derivative of the position with respect to coordinate time (not proper time).
Expressed this way it's obvious that all the entries in the stress-energy tensor have the same dimensions of $ML^2T^{-2}$ (divide by $L^3$ to turn it into a density). So the only remaining question is how the ensemble properties like momentum density and pressure emerge from the point particle description.
The $T^{00}$ element is easy since that's just $\gamma mc^2$, which is the energy. So add up all the point particles and you get the total energy.
The $T^{i0}$ elements look like $\gamma mv^ic$, so add up all the point particles and you get the total momentum multiplied by the velocity in the time direction. Likewise the non-diagonal $T^{ij}$ elements give the total momentum multiplied by the velocity in the $j$ direction. Both are momentum fluxes.
The diagonal elements (other than $T^{00}$) look like a kinetic energy $\gamma m(v^i)^2$. If you consider an ensemble of particles with random velocities (e.g. thermal velocities) then the kinetic energy is simply related to the pressure, and that's why the diagonal terms are effectively a pressure.
One tries to encapsulate "numbers" into tensors because they are related (in a precise way) and it is more compact to write them as a whole object.
This tensor in particular is the result of applying Noether's Theorem to a general space-time transformation, it is in fact the conserved current density that is induced by the space-time symmetry. This means that you have conservation laws that can be expressed in an integral form, and the conserved objects that come from them are called energy, momentum, etc.
The indices or components (the $\mu$ and $\nu$) allow you to find the specific conserved quantities as you said, but you can observe directly where they come from when you look at the definition in terms of derivatives of the Lagrangian; it's no wonder they have those meanings in the end.
Edit: I think this answer gives extra details in the same vein.
Best Answer
First, I suppose here that we are not considering here gravitational fields. (the explanation will be given below). So, we are considering matter fields, radiation fields, etc..
You have first to write a Lagrangian density $\mathcal L_{NG}$ for your problem (here NG stands for non-gravitational). For instance, the electromagnetic field has the Lagrangian density $\mathcal L_{NG} = -\frac{1}{4}g^{\mu\rho} g^{\nu\lambda} F_{\mu\nu} F_{\rho\lambda}$
Then, you may obtain the Hilbert stress-energy tensor :
$$(T_{\mu\nu})_{NG} = \dfrac{-2}{\sqrt{-g}} \dfrac{\partial (\sqrt{-g}\mathcal L_{NG}) }{\partial g^{\mu\nu}} \tag{1}$$
The formula $(1)$ may be justifyed by considering, that, ultimately, the stress-energy tensor for non-gravitational fields, $(T_{\mu\nu})_{NG}$, is the source of gravitation ($g_{\mu\nu}$), so, even, if, in a problem, you are not considering gravitational couplings, you may write a Lagrangian density with an explicit metrics $g^{\mu\nu}$ , and then calculate the Hilbert stress energy-tensor. One avantage is that this stress-energy tensor is automatically symmetric.
Note that, then, the formula $(1)$ does not appy to the lagrangian density of the gravitational fields itself