This is a system of two capacitors in parallel, so the total charge is given by $$ Q_\mathrm{tot} = (C_1 + C_2)V ~.$$ When a dielectric is inserted into capacitor 1, it's capacitance changes: $C_1 \to \kappa\,C_1$. However, the total charge is conserved. Thus $$ Q_\mathrm{tot} = (\kappa\,C_1 + C_2)V' ~. $$ This implies that
\begin{align}
(C_1 + C_2)V &= (\kappa\,C_1 + C_2)V' \\
\Rightarrow V' &= \frac{(C_1 + C_2)V}{\kappa\,C_1 + C_2} \\
\therefore V' &= \frac{Q_\mathrm{tot}}{\kappa\,C_1 + C_2} ~.
\end{align}
If this result is not working wherever you are submitting it, then it may be that the website is wrong, or you are misreading the question.
Consider the three-terminal device that is your stacked capacitor:
A ----=============================================
(dielectric medium ɛ)
=============================================---- B
(dielectric medium ɛ)
C ----=============================================
All three plates have the same area A. It's pretty clear what happens when A is left "floating". (Defining floating: before operation we ground A so that $Q_A = 0$, then we disconnect it so that $A$ is not connected to any particular other component.) When we float A and measure the capacitance between B and C, or when we float C and measure the capacitance between A and B, we get an electric field $E = Q / (\epsilon A)$ and capacitance $C = \epsilon A / d$ where $A$ is the area of these plates, $Q$ is the charge on one terminal, and $d$ is the distance between them. The key observation here is that the capacitor has a very large area compared to its width, so the electric field outside of the capacitor tends to 0, so neither A nor C really "matters" when it's floating in that 0 electric field.
It is somewhat harder to think about what happens when we float B and then measure the capacitance between A and C. The electric field needs to be 0 inside B, because it is a perfect conductor and any electric field will cause current to flow. At the same time, the overall charge is 0 and the situation is still "capacitor-like" ($A$ is much much larger than $2d$ if it's much much larger than $d$) so the electric field outside of the parallel plates should tend to 0. Since the charge on the plate directly creates a discontinuity in the electric field, we have to come to this conclusion: the field in the dielectric between B and C is the same when we put $+Q$ on A, $-Q$ on C, as when we floated A, putting $+Q$ on B and $-Q$ on C. It has to be, because it's the same jump discontinuity from the same starting point ($E = 0$ outside the stack). The same must also be true between A and B. The field must be $Q / (\epsilon A)$ in both dielectrics.
The condition that $E = 0$ within the middle plate means that we induce a surface charge of $+Q$ on the BC side of B, and a surface charge of $-Q$ on the AB side of B. When you include those surface charges, it "looks exactly like" two capacitors in series, and you expect half of the capacitance.
And that's exactly what happens if you ignore B, too! If you ignore B, then you've got a constant field of $E$ over twice the distance, so $V = 2 E d$ for the same $Q$, so it takes twice as much voltage to get the same charge on each plate. So while B is doing "something", it's actually doing the nothingest something it can do. So you're right to intuit that it should just work like a single two-plate capacitor, if you ignore the thickness of the B plate in the calculation of how thick the capacitor is.
Now that we understand this, here comes situation 1. For situation 1, the easiest way to get an analogous result is to connect A and C with a wire, so they are at the same voltage, each plate holding charge $Q/2$ while the B-plate holds charge $-Q$. Then you are correct to intuit that this just looks like two capacitors in parallel. What happens? Well, the field is still 0 outside the system. The charge $Q/2$ therefore means that we have half the electric field inside the AB and BC dielectrics, which means that the same charge requires only half the voltage, so the capacitance doubles.
Now what if, as you say, we put charge +Q on plate A, +Q on plate C, and -Q on plate B? Well, we have a problem: the overall charge is no longer 0. Under the same "parallel plate" assumption that $A$ is much much larger than $d$ we find the fields by the principle of superposition:
E = + Q/(2 A ɛ)
A ----=============================================
E = - Q/(2 A ɛ)
=============================================---- B
E = + Q/(2 A ɛ)
C ----=============================================
E = - Q/(2 A ɛ)
Now, we can't even define capacitance unless we choose two points to measure a voltage between. Suppose you want the points A and C: the voltage between these plates is 0, and the capacitance is infinite. Actually, since this voltage is 0, we wouldn't change the system fundamentally by connecting A to C. So then you can consider the voltage between A and B, and you get the same result as before, twice the capacitance as if C weren't there. The surplus charge on A and C, while it superimposes on the electric fields, doesn't affect the capacitances involved.
Best Answer
I don't have exactly the figures to deal with your specific examples but these ones will be useful. Just replace the second medium with vacuum to deal with your specific example. You can also adjust the argument to have any thickness of each medium.
You are correct that, on the surface of the dielectric, the surface charge density $\sigma_i$ will be less than the surface charge density $\sigma$ on the conducting plates. If there were only one type of material the situation would be like this:
It's not hard to see how this generalizes to two materials. Indeed, the argument about surface charges does not depend on the number of materials, as you have correctly guessed.
But one does not use $\sigma_i$: instead one must know $V_1$ and $V_2$. To get these one needs the electric fields in medium $1$ and $2$, given respectively by $$ E_1=\frac{\sigma}{\epsilon_1}\, ,\qquad E_2=\frac{\sigma}{\epsilon_2} $$ where $\sigma$ is still the surface charge density on the conducting plates. The effect the charges induced in the dielectrics is to reduce the fields $E_1$ and $E_2$, and this is captured by $\epsilon_1$ and $\epsilon_2$. The fields $E_1$ and $E_2$ are the net fields, resulting from the superposition of the field created by the plates and the field from the induced charges with surface density $\sigma_i$. Since the $\vec E_k$ field are constant in each material (by Gauss's law), we have \begin{align} V_1&= E_1\times \frac{d}{2}\, ,\qquad\qquad\qquad\qquad V_2=E_2\times \frac{d}{2}\, ,\\ V&=V_1+V_2=\sigma\frac{d}{2}\left(\frac{1}{\epsilon_1}+\frac{1}{\epsilon_2}\right)\, . \end{align} You can see here again that everything is expressed in terms of the charge density $\sigma$ on the conducting plates; the fact there are bound charges on the dielectrics is included in the values of the permittivities $\epsilon_1$ and $\epsilon_2$.
Since now $C=Q/V$ where $Q=\sigma A$ is the free charge on the plates of area $A$, you find $$ \frac{1}{C}=\frac{V}{Q}=\frac{V_1}{\sigma A}+\frac{V_2}{\sigma A} $$ and the usual relation for capacitors in series. Note again that the same $\sigma$ of free charge appears in computing the net charge on the conducting plates. Thus, the charges on the surface of the dielectrics don't matter in the sense they never enter directly: their effect - which is to lessen the electric field in the materials - is through the introduction of $\epsilon_1$ and $\epsilon_2$.
Sorry I'm a clutz for editing the size of figures...
(Figure credits: Sears and Zemansky's University Physics, by Young and Freedman)