You have the following equations, valid everywhere
\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{D}=\rho, \quad \vec{D}=\epsilon \vec{E}.
In your case you have
\epsilon=\Theta(z)+\epsilon_{r} \Theta(-z).
To get the boundary conditions you just apply Stokes's theorem to the first equation with an arbitrary rectangular surface perpendicular to the surface. This leads to the condition that the tangential components of the electric field must be continuous across the surface, i.e., in your case
\vec{e}_z \times [\vec{E}(x,y,0^+)-\vec{E}(x,y,0^-)]=0,
because \vec{e}_z is the normal vector of the dielectric's boundary surface.
Further you use an infinitesimal cube parallel to the boundary on the 2nd equation. This yields the boundary condition
\vec{e}_z [\vec{D}(x,y,0^+)-\vec{D}(x,y,0^-)] = \sigma.
Here \sigma is the surface charge density on the surface of the dielectric.