$\let\eps=\varepsilon \let\rho=\varrho \let\sig=\sigma \def\vD{\vec D}
\def\vE{\vec E} \def\ns#1#2{#1_{\rm#2}} \def\cE{{\cal E}} \def\cEa{\ns\cE a} \def\cEb{\ns\cE b} $
Given that @probably_someone choose to answer with no reference to displacement $\vec D$, I will take the alternate way. I agree that using polarization vector a deeper understanding is got as to what happens in the dielectric. But if electric displacement still survives, almost 150 years after Maxwell, there are good reasons: using it in problems with dielectrics is generally much simpler.
Now, how is the charge distributed?
If the plates are much larger than their distance, over most of the
surface (borders excepted) $\sig$ is constant.
Let me summarize the relevant equations. In electrostatics they are
three: two Maxwell's equations
$$\nabla \times \vE = 0$$
$$\nabla \cdot \vD = \rho$$
and one constitutive relation, which for isotropic dieletrics is
$$\vD = \eps \vE$$
with $\eps$ a scalar quantity. I assume your dielectrics are homogeneous, so that only two values are needed. You will excuse me if I prefer $\eps_1$, $\eps_2$ to your $K_1$, $K_2$. AFAIK, $\eps$ is a universal symbol for dielectric constant.
Given the problem's symmetry, it is obvious that both $\vE$ and $\vD$ are wherever perpendicular to the plates, and are uniform within each dielectric separately. They could be different between the upper and the lower dielectric, but use of Gauss' theorem for a small cylinder put across the interface shows that $\vD$ stays the same, as there is no free charge there. (Free charges are located on plates only.)
Continuing with $\vD$, we can see its advantages: we can reason when dielectrics are present exactly like we do in vacuum. Between the plates $\vD$ is directed from positive to negative plate and its modulus equals $\sig$. Moreover, $\vD=0$ outside (always neglecting border effects).
How can we find the surface charge densities at the interface of the
two dielectrics?
This cannot be answered using $\vD$ alone, as it ignores bound charges. But once you know $\vD$, $\vE$ follows immediately: $E_1=D/\eps_1$, $E_2=D/\eps_2$. The same cylinder we used before tells us that
$$\sig' = \eps_0 (E_2 - E_1) = \eps_0\,D \left(\!{1 \over \eps_2} -
{1 \over \eps_1}\!\right) = \eps_0\,\sig \left(\!{1 \over \eps_2} -
{1 \over \eps_1}\!\right).$$
Qualitatively, from what I understand, the energy stored between the plates should increase. How do we find the energy stored in each dielectric slab?
This is an interesting question. I can't see why energy should increase, in your opinion. Actually the opposite happens. Let me recall the general expression for energy density within a dielectric:
$$U = \frac12\,\vE \cdot \vD.$$
Then total energy before dielectric insertion is
$$\cEb = \frac1{2 \eps_0}\,\sig^2 S\,(d_1 + d_2)$$
where $S$ is plate's area. After insertion we have
$$\cEa = \frac1{2 \eps_1}\,\sig^2 S\,d_1 + \frac1{2 \eps_2}\,\sig^2 S\,d_2 =
\frac12\,\sig^2 S \left(\!{d_1 \over \eps_1} +
{d_2 \over \eps_2}\!\right)\!.$$
Since both $\eps_1$ and $\eps_2$ are greater than $\eps_0$, it's clear
that
$$\cEa < \cEb.$$
As a matter of fact, it is known that dielectrics are pulled inside electric fields. In other words, by inserting a dielectric in a capacitor useful work can be obtained.
