If we short two terminals of a cap, will the charge just spread out between them or will it oscillate back 'n forth before it spreads out?

Assuming the cap (capacitor, I assume) has some charge on it?

the short will cause a high current as the charge tries to equalize. The current is limited by the wire resistance and inductance, and the resistance and inductance internal to the cap. The current will oscillate back and forth at the resonant frequency of the inductance and capacitance for a few cycles until all the energy is dissipated in heat and RF radiation. At that point there will be no charge on the cap, both plates will be at the same potential.

Not clear what you mean by the charge "spread out" ?

No, just its plate has negative charge ( contains electrons ) goes through the short to reach other plate has positive charge. Its charged energy turns into heat along the path in a single action.

Charge can not be spread out between two capacitor plates that are shorted together.

The self induced current from the collapsing field of the capacitor's series inductance will initially oppose and finally aid the capacitor's external current from one capacitor plate to the other when the capacitor's terminals are shorted. Most all of the electrical energy initially stored in the capacitor's dielectric and the inductor's magnetic field will be dissipated in the circuit's resistance as the two energy sources are now series aiding. A capacitor whose terminals are shorted can not be charged in the opposite direction therefore no oscillation will occur. Maybe current damping but I don't think the current will ever reverse directions because the capacitor can not be charged in the opposite direction with it's terminal's shorted. Therefore no oscillations and all the stored electrical energy will have been dissipated by the time the current quits flowing in the initial direction.

Note: This is my opinion and may or may not be valid so don't bet your bottom dollar on it.