So, DMSO (dimethyl sulfoxide) can form metal complexes in both $\ce{\kappa-O}$ and $\ce{\kappa-S}$ mode i.e. binding with the oxygen or the sulfur respectively. The general explanation given is that softer metal ions bind to the soft centre $\ce{S}$ and hard metal ions bind to the hard centre $\ce{O}$.
From the IR data it can be inferred that the binding of DMSO to a metal ion via $\ce{O}$ causes the $\ce{S=O}$ bond stretching frequency to go down. For example the $\ce{S=O}$ stretch in free DMSO (liquid) is $\pu{1055 cm-1}$, whereas the stretch in $\ce{[Co(dmso)6]+}$ appears at $\pu{950 cm-1}$. This makes sense to me because the oxygen can only act as $\pi$ donor (apart from the usual $\sigma$ donation), and donate electrons to the metal $\mathrm{t_{2g}}$ from bonding $\pi$ orbitals, which will weaken the bond between $\ce{S}$ and $\ce{O}$.
However, when DMSO binds to a metal ion via $\ce{S}$, the $\ce{S=O}$ bond stretching frequency goes up. For example, in $\ce{PdCl2(dmso)2}$ shows the $\ce{S=O}$ stretch at $\pu{1116 cm-1}$.
But how is this possible? The sulfur is already electron deficient due to the strongly electronegative $\ce{O}$, and should be a $\pi$-acceptor, and accept electrons from metal into the $\pi^*$ orbitals. This means that the $\ce{S=O}$ bond should also be weakened in this case. But this does not happen. Why? Does it have something to do with the molecular orbitals of DMSO?
[Note: According to this post, the $\ce{O}$ is mainly $\pi$-donor and the $\ce{S}$ $\pi$-acceptor for DMSO, and I agree with the argument in that answer. But it does not say anything about the strengthening of the $\ce{S=O}$ bond in $\ce{S}$ bonded complexes.]
The pictures of HOMO and LUMO of DMSO respectively: