Does Compton effect take place because at energies higher than required for photoelectric effect, the electron would be destabilized after absorbing a photon of such a large amount of energy and therefore dissipates some reproducing another photon and using some kinetic energy itself to be scattered?
Both the electron and the photon are elementary particles, and the interactions of elementary particles need quantum mechanics. Feynman diagrams are an iconal representation of the calculations necessary to get measurable predictions for the interactions of elementary particles. In this case Compton scattering:
Elementary particles interact at a point, they are point particles. Here there are two diagrams contributing to lowest order, a real photon hitting a real electron as input, a real photon and a real electron as output. In between there exists a virtual electron, within an integral. It is called "virtual" because it is off mass shell.
There is no "destabilization", unless one means the off mass shell propagating "electron", with the quantum numbers of the electron but not its mass. Calculating the diagrams will give the probability distributions for the incoming photon to transmit energy to the exiting electron and by momentum and energy conservation the outgoing photon will have a lower energy. ( There exists inverse Compton effect , where a low energy photon gains energy but it is a state relevant to astrophysics)