Platinum Orbitals in Cisplatin Forming Bonds with LCAOs of Chlorine
I figured it out!

We have to start by looking at d orbital shapes:



To form bonds, the d orbitals of platinum must have the same symmetry (i.e. shape) as the molecular orbital that's approximated by the LCAO. 

We start with the bonding orbital, which looks like this: (sorry about the pencil-and-paper phone picture). 



We can see we need a d orbital with exactly one lobe in the zx plane (x is pointing to the right in the above). The only d orbital that fits the requirement is the $\textrm{dz}^{2}$.

Moving on  to the anti-bonding orbital, which I can't post because I don't have at least 10 reputation, we see that it has four lobes with two nodal planes. It looks like a $\textrm{d}_{xz}$ orbital, in fact.

We can see we need a d orbital in the zx plane that has two lobes. Clearly, this is the $\textrm{d}_{xz}$ orbital.

Easier than I thought once I got the concepts down!