When you say that the surface charge density is given by the normal component of the electric field, you are assuming the field is zero outside the surface. Similarly if you take the potential itself to be zero outside the surface, then the other term can be interpreted as the electric dipole layer necessary to give the discontinuity in potential required by your boundary conditions.
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You can understand the dipole layer by looking at a plane at $z=d/2$ with surface charge given by charge per area $\sigma$, above a second plane at $z=-d/2$ with charge per area $-\sigma$. The electric field will $-\sigma\hat z$ for $|z|<d/2$. The potential difference between $z = d/2$ and $z=-d/2$ will be $\sigma d$. Taking the limit that $d\rightarrow 0$, $\sigma\rightarrow \infty$ with $\sigma d$ fixed will give an elementary dipole layer with dipole moment per unit area of $\sigma d$, and this same discontinuity in the potential. Getting infinitesimally close to the surface, a local surface dipole moment density looks like this plane of surface dipole moment.