Determining the oxidation state of a bridging platinum complex
For organometallic complexes it's often easier to throw the "more electronegative element" idea out entirely and focus on the characteristics of the ligands. This is what @andselisk's comment is referring to. Generally ligands are partitioned into two types, $\ce{L}$ and $\ce{X}$:


$\ce{L}$ refers to neutral ligands: for example $\ce{CO}$, phosphines, $\ce{H2O}$, alkenes (via the pi bond), ... and $\ce{SMe2}$ falls comfortably in this category.
$\ce{X}$ refers to anionic ligands: for example halides, hydroxide, azide, alkyl anions such as $\ce{Me}$, ...


Other ligands can be constructed as sums of these: so for example η4-butadiene would be $\ce{L2}$ (two pi bonds coordinating to the metal), an allyl group would be $\ce{LX}$ (one pi bond plus one anionic centre - ignore resonance), and so on.

The oxidation state of the metal, then, is simply the charge on the metal plus the number of X ligands. In this case each Pt centre is neutral, and has two L and two X ligands. That makes for a total oxidation state of +2.

As another example, the platinum in Zeise's salt has one L ligand (ethylene), three X ligands (chloride), and a negative charge: so its oxidation state is $-1 + 3 = +2$.

Of course, this method will fall apart very quickly when there are non-innocent ligands: $\ce{NO}$ and $\ce{O2}$ are the examples that come to mind, although there are many more. But for relatively simple complexes they suffice.