Second order Jahn-Teller effect for d0 complexes
After quite some research I finally found the original work on the problem. This comes from: 
S. K. Kang, H. Tang, T. A. Albright, J. Am. Chem. Soc. 1993 115 1971 - 1981. 

I won't discuss the symmetry parts, simply because I have no clue about symmetry and point groups and transformations. I copied the original description for the necessary relationships for a second order JT-effect to happen in the question, so if you know more about this you can probably confirm that.

So to my understanding what is happening for d0 compleces is the following. You really get a mixing of the lower and filled $\mathrm{t_{1u}}$ Orbital with the higher unoccupied $\mathrm{t_{2g}}$ orbital. Both of them split in a set of one orbital that is unchanged in energy, $\mathrm{a_{2}''}$ (for the original $\mathrm{t_{1u}}$) and an $\mathrm{a_{1}'}$ (for the original $\mathrm{t_{2g}}$), and two degenerate orbitals with $\mathrm{e'}$ symmetry that is stabilized and therefore lowered in energy for the original $\mathrm{t_{1u}}$ and elevated for the original $\mathrm{t_{2g}}$. According to the paper this is done by a distortion of the octahedron by a normal mode of $\mathrm{t_{2u}}$ or $\mathrm{t_{1u}}$ symmetry. The first is said to equal a rotation along a threefold axis in the octahedron to get a trigonal prism and change the $\mathrm{O_{h}}$ symmetry to $\mathrm{D_{3h}}$.

You can also see this in the appendix of the book:
T. A. Albright, J. K. Burdett, M.-H. Whangbo, Orbital interactions in chemistry, second edition, Wiley, 2013.



The next question is, why does this stop at the trigonal prism and not continue in mixing the other states as well? And the answer is, for some hypothetical hydrides this seems to happen. While in a d2 complex the new $\mathrm{a_{1}'}$ is filled and therefore an elevation in energy quite unvavorable a d0 complex should further distort. Then $\mathrm{a_{1}'}$ and $\mathrm{a_{2}''}$ mix and distort along a normal mode of $\mathrm{a_{2}''}$ symmetry that changes the trigonal prism $\mathrm{D_{3h}}$ to something they call pyramidalization with $\mathrm{C_{3v}}$ symmetry. This step can also be achieved directly from the octahedron by mixing all orbitals at once and twisting a triangular face of the octahedron.



Now the final question was, why do we observe so many trigonal prisms then? And their final remark on this (unless I skipped something) that it's about the ligands. What ligands favor a second order Jahn-Teller effect for a d0 metal center? Basically only strong σ-donors. They mustn't have any π-donating effects. Since the HOMO LUMO gap has to be quite narrow the $\mathrm{t_{1u}}$ orbital has to be high. So no electronegative ligands. But also the $\mathrm{t_{2g}}$ orbitals have to be low, so a strong ligand splitting is favored but in a way that the $\mathrm{t_{2g}}$ is lowered. And most probably a strong nephelauxetic effect is advisable as well. This leads to alkyl and hydride as possible candidates. For the hydride many of the desired complexes are just unknown but the calculations show that is could distort to $\mathrm{C_{3v}}$. For the alkyl like many prismatic organometallic alkyl complexes we know the methyl seems to be to bulky to stabilize the pyramidal structure.

In the same sentence they also mention that for example alkoxides are not forming prisms. So this leaves me with the question how Mo-S or W-O structures in solid state still show the trigonal prism. But that has probably different reasons so I'll exclude them here.