The problem with your idea (S having more protons, thus having a greater force on the electrons) is this: While there are more protons in the nucleus of S, there are additional electron shells (with electrons in them). This does two things, first off, there are more electrons repelling lone pair electrons or electrons from other atoms. This, in and of itself, does not counter the additional force from the increased charge in the nucleus, as can be seen by Fluorine having a higher electronegativity than Oxygen. Oxygen has the same number of electron shells as Fluorine, just one more electron and one more proton yet Fluorine has a stronger attraction towards taking on additional electrons. With that example we see that a 1:1 increase in protons and electrons increases the atom's ability to pull in electrons, as long as there are the same number of electron shells. However, the additional shells increase the distance between the atoms, as shown by Coulomb's law (Force = q1*q2/(4*pi*epislon(not)*r^2)) where q1 and q2 are the charges on the charged particles, epsilon(not) being the permittivity of free space constant, and r being the distance between the charges squared. As you can see, force decreases on the order of the inverse of the distance squared. (For instance, if the force were initially Fi at distance r, at distance 2r it becomes Fi/4). This increased distance causes the electronegativity to decreases as you go down a family on the periodic table of elements. As such O actually has a stronger pull on electrons than S would, since the electrons are closer to the nucleus with O.
As far as your question goes, why SF2 has larger bond angles than those of OF2, from what I looked up it doesn't. I found very similar numbers to what Schrödinger found, so I would assume those are roughly accurate. I'm not sure why this is the case, though, so I hope you figured it out from someone else. If anyone knows why these numbers are the way they are, I'd be very interested in hearing.