Why does the polarization type ($s$ or $p$) of electromagnetic waves effect the reflectivity and reflectivity of the light differently

Well, I encourage you to work through the derivation of the Fresnel Equations. What you’ll find is that the difference between $s$ and $p$ polarizations comes from the boundary conditions. Due to Gauss’ Law, the parallel component of the electric field is continuous across the boundary, while the normal component is discontinuous. The $p$ polarization has a component of field normal to the interface, while $s$ does not.