The interpretation is that a discontinuity in E is a surface charge density, which is not a charge density per unit volume, but a charge per unit area. A smooth charge density per unit volume keeps E smooth, but if you have a plate of surface charge it has an infinite charge density per volume and make E jump. Such infinities can only appear on a physical boundary between two phases, and must be confined to the surface. They happen in a conductor at voltage.
Surface currents of finite amount of current per unit give rise to a discontinuity in B parallel to the plate, as you can see from Ampere's law (or by superposing many infinitely thin current wires next to each other). The B field from a surface j is a discontiuity in B parallel to the surface, and any bulk current is a current density, and keeps B smoothly varying.
Both results are correct. As Luboš Motl pointed out in his comment, we can get the geometrical optics approach answer from the wave method answer by averaging it over 1 full period.
Perhaps you made a mistake somewhere when averaging.
If we calculate the average carefully:
$\bar{T}=\frac{1}{2\pi}\int _{-\pi}^{\pi}T d\varphi$
$\bar{T}=\frac{1}{2\pi} \times T_{01}T_{12} \int_{-\pi}^{\pi}\frac{1}{(r_{10}r_{12})^2-2r_{10}r_{12}cos\varphi+1}d\varphi$
we'll get this
$\bar{T}=\frac{1}{2\pi} \times T_{01}T_{12}\left|\frac{2tan^{-1}\left(\frac{1+r_{10}r_{12}}{1-r_{10}r_{12}}tan{\frac{\varphi}{2}}\right)}{1-(r_{10}r_{12})^2}\right|_{-\pi}^{\pi} $
you can check here
now it's not hard to see that
$\bar{T}=\frac{1}{2\pi} \times T_{01}T_{12}\frac{(2\pi)}{1-R_{10}R_{12}}=\frac{T_{01}T_{12}}{1-R_{10}R_{12}}$
same as the one obtained from geometric optics method
EDIT(to respond to the EDIT part of the question):
First of all, let's make everything clear so that the equation that you quoted from Wikipedia make sense. The amplitude here means the magnitude of electric field. The $t$ simply means the ratio of transmitted and incoming electric field (We were using different definition of $t$ in the calculations above, the $t$ that we use above includes other factors beside the ratio of electric fields). And the transmittance $T$ here represents the fraction of power transmitted to the medium 2. Here the power $P$ is proportional to
$P\propto IA$
where the intensity is proportional to $I\propto nE^2$. And the beam area is proportional to $A\propto cos\theta$, it's because the beam cross section gets smaller as it bends toward the boundary plane (see this image). Putting them together we have
$P\propto nE^2cos\theta$
so from the new definition of transmittance $T$ we get
$T=\frac{n_2 cos\theta_t}{n_1 cos\theta_i}t^2$
and since the incoming and reflected rays have the same $cos\theta$ and $n$, the reflectance is simple
$R=r^2$
we have calculated that
$\bar{t^2}=\frac{t_{01}^2t_{12}^2 }{1-R_{10}R_{12}}$
multiply both sides with $\frac{n_2 cos\theta_t}{n_0 cos\theta_i}$
$\bar{T}=\frac{t_{01}^2t_{12}^2 }{1-R_{10}R_{12}}\frac{n_2 cos\theta_t}{n_0 cos\theta_i}$
We can check
$T_{01}T_{12}=\left(\frac{n_1 cos\theta_m}{n_0 cos\theta_i}t_{01}^2\right)\left(\frac{n_2 cos\theta_t}{n_1 cos\theta_m}t_{12}^2\right)=\frac{n_2 cos\theta_t}{n_0cos\theta_i}t_{01}^2t_{12}^2$
Thus again we have
$\bar{T}=\frac{T_{01}T_{12}}{1-R_{10}R_{12}}$
Best Answer
I’m not sure which derivation you’re looking at, but (a) the tangential component of $D$ is virtually always continuous (absent a static charge density which simply contributes a static field), and (b) including a current density poses no problem. Take a look at Ampere’s Law (time-harmonic approximation): $$ \nabla\times H = J + i \omega D $$ It’s clear that the free current and displacement current terms both contribute to the curl of H, they just have a relative phase shift: $$ \nabla\times H = (\sigma + i \omega \epsilon)E $$ So we can bundle $\sigma$ (the conductivity, not the surface charge density) into $\epsilon$, giving it a complex value, but nothing else about Ampere’s Law changes. Therefore, the same derivation for the Fresnel coefficients applies in the case of a complex $\epsilon$.
Just in case you don’t believe your book, trust that the validity of the Fresnel coefficients for complex-valued $\epsilon$ has been empirically verified countless times over decades. I have used it myself to calculate the transmission spectra of thin metal films, which I used to successfully model my data.