What is the mechanism behind the decomposition of enediol phosphate intermediate to methyl glyoxal?
This reaction can occur only when the $\ce{C-O}$ bond to the phosphate group
  lies in a plane that is nearly perpendicular to that of the enediol so
  as to permit the formation of a double bond in the intermediate enol
  product.


In solution, the enediol intermediate readily breaks down with the elimination of the phosphate at C3 to form the toxic compound methylglyoxal:



The spontaneous decomposition of the enediol intermediate in the triose phosphate isomerase (TIM) reaction to form methylglyoxal through the elimination of a phosphate group. 

In order for this elimination to occur, the bond to the phosphate group must lie, as shown above (a), in the plane perpendicular to that of the enediol.

This is because, if the phosphate group were to be eliminated while this bond was in the plane of the enediol as shown in (b), the $\ce{CH2}$ group of the resulting enol product would be twisted 90° out of the plane of the rest of the molecule. Such a conformation is energetically prohibitive because it prevents the formation of the enol’s double bond by eliminating the overlap between its component $\mathrm{p}$ orbitals.

Reference 

Voet and Voet Biochemistry