Stacking batteries and electron flow About Fuel Cells and Batteries:
I understand the analogue with increased water height as two cells (battery or fuel) are connected in series.
However if we for instance have two batteries contected in series cell A and B. 
They both has a - and + terminal.
+A (terminal) is connected to -B.
+B is connected through a load to -A.
Electrons (1) will then flow from -A through load to B+, since there is a potential difference between -A and B+ terminals.
Similar electrons (2) will flow from -B to +A.
However are these electrons (1 and 2) the same electrons?
I mean since electrons that are produced in a oxidation proces at terminal -, flows to terminal +, and then used in a redox process at terminal + cannot continue its flow through the "stack" of cells, right?
If the electron flow 1 and 2 are not the same electrons then how come that the potential is raised?
I mean, are the electrons just "representing" the flow, but they are not the actual cause of the flow.
Similar to heat gradient from room A (with air particles P) to room B. If the temperature is higher in room A then P particles will flow to room B, but even though particles P perhaps are stopped by some membran in between room A and B, the heat gradient is still present
 A: The straightforward answer to your question is no, they are not the same electrons.
To understand why requires us to give some deeper consideration to what is going on chemically within the two cells.  I am going to give my explanation using the simplest chemistry I can imagine, which is a Zinc-Copper battery.
The Zinc-Copper cell is a rod of zinc and a rod of copper, wired together via a load such as a resistor with parallel voltmeter ( this can be used as an a current meter).  Then placing the zinc in a zinc salt solution, eg zinc chloride, and the copper stick in a copper salt solution, gives two so called half-cells, and linking the two with a salt bridge (this is just a link that allows the ions to move, it can be wet paper, or a solution filled tube) forms the battery.  
The key is to understand what happens inside the half cells:  The copper and zinc will tend to dissociate in the solution to produce electrons and ions e.g $\mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + \mathrm{2e^-}$.  The electrons can then flow around the conductor and reach the other cell.  In order to provide charge balance ions will take the opposite journey, through the salt bridge and reach the other cell to undergo a redox reaction in that cell.
The direction of flow of the electrons and ions is determined by the "standard reduction potentials" of the two half cells, this is in some sense a measure of the amount of energy needed to liberate the electrons from the positive ion.  The half cell with the greater reduction potential will be the anode, and the smaller reduction potential will be the cathode, as a greater reduction potential means it is easier for the half-cell to accept electrons, and a lower one easier to donate electrons, (you may recall OILRIG from chemistry lessons).
In my example above we would find that the copper half cell reaction has a reduction potential of $E^{\circ}=0.337$ and that the Zinc half cell has a reduction potential $E^{\circ}= -0.763$, (this may also be expressed as the oxidation potential $E^* = -E^{\circ}$).  Thus the Zinc half-cell will be the cathode in this hypothetical battery.  Inside the cells therefore, a redox reaction will be occurring in which zinc metal is oxidised at the cathode, liberating electrons to flow through the load, and simultaneously at the anode copper ions are reduced to copper metal providing a sink for the electrons arriving at the other end. 
Now to the second part of your question, the key is that the electrons move due to the electric fields caused by potential differences. 
It is important to understand that potentials are relative.  The concept of an absolute potential does not exist in physics, and we are always concerned with the "potential difference" between two points.  This comes from the expression for the potential $\vec{E} = -\nabla\Phi$, which can be simplified in the case of two point potentials such as the terminals of a battery to $E = \Delta V/l$ where the $\Delta V$ is the potential difference across the battery and $l$ the length of the battery.
For example the potential difference across two terminals of a battery is due to the oxidation and reduction potentials of the two terminals as discussed above.  This leads to a measurable potential difference ($V$) on your voltmeter, but also means that there is an electric field inside that battery due to that potential difference.  Now, if we take two of those cells and stack them one on top of the other, and measure the new potential difference ($V^*$) we will see double the potential difference we had before.  You can understand this by seeing that there will be twice the length now which contains the electric field, and as such if the electric field strength is the same, but twice as long in distance now, we must have twice the electric potential difference across the two cells as across a single one:
\begin{equation}
E = \frac{\Delta V^*}{2l} = \frac{\Delta V}{l} \quad \rightarrow V^* = 2V
\end{equation}
Refs:
http://www.science.uwaterloo.ca/~cchieh/cact/c123/emf.html
http://en.wikipedia.org/wiki/Electrostatics#Electrostatic_potential
A: Not immediately, it takes a while. Even with just a simple bit of wire, the electrons flowing in are not the electrons flowing out on the other end. You have to understand there are a lot of metal atoms in the wire itself. When you push electrons into the wire, this displaces the electrons which are already there. This is just the same as with a filled water pipe: when you pour in more water on one side, that new water won't overtake the water. The first water to come out on the other side of the pipe will be the water which already was present in the pipe.
