Why does the d orbital size decreases on addition of electrons?
There is not much to explain because the concept is wrong. There is no sulphur compounds in which the 3d orbitals (remember: they have an energy level similar to 4s) are involved in bonding in any noteworthy way, shape or form.

From what I believe I may remember about the outdated concept, ‘one d orbital’ refers to cases like $\ce{SO3^2-}$ with sulphur in the $\mathrm{+IV}$ oxidation state while ‘two d orbitals’ would refer to e.g. $\ce{SO4^2-}$ or an oxidation state of $\mathrm{+VI}$. In the Lewis formalism, the answer is easily deduced if you draw octet-abiding structures: $\ce{SO3^2-}$ will include a formal single positive charge on sulphur while in $\ce{SO4^2-}$ sulphur has a formal double positive charge. The higher an oxidation state, the more contacted an atom becomes.

A more correct answer would elaborate on ($\mathrm{sp}^n$-type, formed only by sulphur’s s and p atomic orbitals) molecular orbitals and the increasing number of electronegative substituents that need to bond to the central sulphur, reducing its charge density and leading to the same effect.