Quantitative contribution of kinetic and potential energies to the binding energy of the $\sigma$ orbital in $\text{H}_2$ or $\text{H}_2^+$ When a hydrogen molecule forms, 4.52 eV of energy is released, while for $\text{H}_2^+$ the binding energy is 2.77 eV. Such a binding energy is the difference of energies that have four terms in them: (1) the kinetic energy of the electron(s), (2) the potential energy of the electron(s) interacting with the nuclei, (3) the electron-electron interaction, and (4) the proton-proton interaction.
Explanations of $\sigma$ bonding in freshman chemistry texts tend to focus on #2. However, if we want to explain the difference in energy between bonding and anti-bonding orbitals, then it seems plausible that there should be a large difference in kinetic energy, #1. This is because the KE of the bonding orbital is low compared to that of the antibonding orbital because the bonding orbital basically a particle in a long box with wavelength in the long direction equal to twice the length of the box. In the antibonding case, this component of the wave-vector should be basically doubled. 
All four of these energy terms can be represented by quantum-mechanical observables, and they can therefore be defined numerically, and calculated for a given set of trial wavefunctions. How much of the truth is captured by explanations that only mention #2?
 A: This self-answer contains the best information I was able to find. The most detailed information I could find on binding energies was for $\text{H}_2$ (Kari 1973), while my only information on bonding versus anti-bonding is for $\text{H}_2^+$ (Schmidtke 1970).
Schmidtke set out to answer almost exactly the question posed here:

In his review on "The Physical Nature of the Chemical Bond" RUEDENBERG concludes that chemical bonding in terms of potential and kinetic energy
  change is not correctly understood unless the validity
  of the virial theorem is preserved The seemingly
  incompatibility of the opinions that bonding is
  either due to a decrease of the kinetic energy (HELLMANN 2) or is originated from a drop of potential
  energy (see e. g. PITZER 3) could be clarified by a
  closer analysis of various energy contributions upon
  molecule formation. 

Here is his graph of the results of his model for $\text{H}_2^+$:

The horizontal axis is the internuclear distance in units of the Bohr radius, while the vertical axis shows the energies in Hartree units, 1 a.u.=27.2 eV. The notation is b for bonding, a for anti-bonding, T for kinetic energy, and V for potential energy. At a distance equal to the equilibrium distance for the bound molecule, the difference in T is about 0.2 a.u., and the difference in V is also about 0.2 a.u., so in this sense the antibonding effect is about equally due to the contributions of kinetic and potential energies. At slightly larger internuclear distances, the V graphs actually cross one another, so that at these distances, the antibonding effect is almost entirely kinetic.
For binding energies, we're comparing one equilibrium state with another equilibrium state, so these energies become direct physical observables, and the calculations in Kari are probably easier to interpret, although they seem to agree with Schmidtke's results. For the differences between the bound state and the state where the atoms are separated, Kari has:
$\Delta V_\text{nuc}=+0.714$ a.u.
$\Delta V_\text{el}=-1.063$ a.u.
$\Delta T_\text{el}=+0.175$ a.u.
So here it seems to be somewhat more accurate to use the commonly encountered explanation that binding is mainly an effect due to $\Delta V_\text{el}$. However, I don't think it's accurate at all when people explain the bonding-antibonding effect as if it were entirely due to $\Delta V_\text{el}$. I think they do this because it avoids the explicit use of any quantum mechanics via the de Broglie relation or the Schrodinger equation, but that's just wrong.
It also seems that the small value of $\Delta T_\text{el}$ is due to some kind of very delicate cancellation in the case of $\text{H}_2$ and $\text{H}_2^+$. On the one hand, we expect if the pair of nuclei acts sort of like $Z=2$, then all energies should scale up by a factor of four, and to the extent that the virial approximation is valid, this would also mean that all kinetic energies should go up by a factor of 4. On the other hand, the particle-in-a-box argument will tend to reduce the kinetic energy of the bonding state. In the approximation that this is a rectangular box with sides $L\times L\times 2L$, doubling the length of one side should reduce all energies by $2.25/3$. By this crude estimate, the kinetic energy would then increase by a factor of about three. In reality, it stays almost exactly the same, so the two effects seem to almost exactly cancel. I wonder whether this still holds at all in other bonds besides the H-H bond.
References
Kari, "Complete variational treatment of the hydrogen molecule: I. The full electronic Hamiltonian," Can. J. Chem 51 (1973) 2055, https://www.nrcresearchpress.com/doi/pdf/10.1139/v73-306
Schmidtke, "Kinetic and potential energy partitioning for antibonding molecular orbitals," Z. Naturforsch 25a (1970) 542, http://zfn.mpdl.mpg.de/data/Reihe_A/25/ZNA-1970-25a-0542.pdf
