At an atomic level, what happens when you connect two batteries in series so that their voltages are added? I can't for the life of me figure this out. I feel like i'm missing some crucial detail about how batteries work.
Imagine two batteries connected in series, like this:
Circuit <= -(Battery A)+ <= -(Battery B)+ <= Circuit

As far as I've studied, this is what happens:


*

*The negative side of Battery B has a surplus of electrons, while the positive side of Battery A has a "normal" concentration of electrons. Since electrons repel each other, they kinda push themselves to Battery A. (I am imagining a tube with one side filled with "elastic" balls tightly packed and the other side with elastic balls but not pressed together. The balls get pushed until they are all at an equal "pressure").

*A number of electrons constantly reach the positive end of Battery A with a "force" of x Volts. In practical terms, the electrons are colliding with the battery with a higher intensity. For some reason, this causes the electrons coming out of the negative end of Battery A to have a "force" of 2x Volts. I don't understand why.
Is it because there is now two times more free electrons at the negative end of Battery A? And thus the difference between this end and Battery B is two times the normal?


*The electrons get pushed again to the positive end of Battery B. From what I understand, the material in the battery "absorbs" them and releases a free electron on the negative end of Battery B.


Please avoid using water analogies. I specifically want to know what happens at the atomic, per-electron level. 
I don't care much about the differences between each battery. If possible, pick a battery type you prefer or just talk about batteries in general (characteristics shared by all batteries).
 A: One way to think of this is as follows: for your two-battery system with the batteries in series, the positive terminal of battery "A" doesn't know or care what voltage the negative terminal of battery "B" is presenting to it. All battery "A" is going to do is pull electrons out of its positive terminal, lift them to a potential that is N volts greater than this (where N is the battery's rated voltage) and present them to its negative terminal. In practice, this means that the voltages of batteries in series are linearly additive.
A: I think that your analogy of "surplus" of electrons is wrong in one sense, and incomplete in another sense. First, what do you consider a "normal" and a "surplus" of electrons? Second, if you have a "surplus" in one side, in the other you don't have a normal, you have a deficit.
Think in a classic Lead-Acid battery. When fully charged you have lead in the negative plate, and lead dioxide in the positive one. In both plates at start you have the usual amount of electrons (82 e per Pb atom in the negative, and 98 per PbO2 molecule).
But they are submerged in a aqueous solution of sulfuric acid, given the possibility to the following redox reaction:


*

*At anode (negative): Pb (s) + HSO4− (aq) → PbSO4 (s) + H+ (aq) + 2 e−

*At cathode (possitive): PbO2 (s) + HSO4− (aq) + 3 H+ (aq) + 2 e− → PbSO4 (s) + 2 H2O (l)


As you can see, for this hemireacctions occur, you have to do take the electrons in the anode and put an equivalent amount in the cathode. It does not work if you do just one, because you need equilibrate charge and matter: An H2SO40 transforms into H+ and HSO4-, you can't have one without the the other (nor have positive without negative, neither transform a complete molecule 98 g of H2SO4 into 1 g of H+.
The voltage is measure of potential job diference. If you have 2 volts between cell electrodes means that the cell is capable of do a 2 J job moving a 1 C negative charge from the anode to the cathode. If you put two cells in series, the job, each one with 2 v, among both are able to do a 4 J job moving a 1 C negative charge from the free anode of one to the free cathode of the other.
Edit 3/8: Someone asks "So if you connect two, do you get a neutral state in the middle?". Initially, no.
Let's start disconnected from the circuit. Electrons can't flow if the hemireaction does not happens. The hemireaction can't happen on its own, without its counterpart. And the counterpart can't happen without its own electron flow.
So, when connected, the carges are going to start neutralising as the electrons flow through the circuit and both hemireactions happen. And that neutralisation is going to happen in all four electrodes, discharging the batteries; because some hemireaction happens in all four electrodes. When the neutralisation is complete, the batteries are completely discharged.
A: Consider first two charged capacitors which are connected in series.  

The potential of plate $A$ relative to plate $B$ is $V_{\rm AB}$ and the potential of plate $C$ relative to plate $D$ is $V_{\rm CD}$ so the potential of plate $A$ relative to plate $D$ is $V_{\rm AB} + V_{\rm CD}$.  
Within each capacitor there is an electric field and the work done by an external force in taking unit positive charge from plate $B$ to plate $A$ is $\int_B^A E_{\rm AB} \,d s$ and from  plate $D$ to plate $C$ is $\int_D^C E_{\rm CD} \,d s$.
The total work done is $D$ to plate $\int_B^A E_{\rm AB} \,d s + \int_D^CE_{\rm CD} \,d s$.
This is the same result as the addition of potential differences.  
What about the conductor between plates $C$ and $D$ as it might appear that the negative charges on plate $B$ should neutralise the positive charges on plate $C$ and due to those charges is there an electric field in the conductor between plates $B$ and $C$.
Well there is no electric field within that conductor and the charges on plates $B$ and $C$ are held in place by the charges on plates $A$ and $D$ respectively.  
So what about ideal cells connected in series?  

If effect cells have two plates which include the terminals.
These terminals/plates get charged up by an electrochemical process which can be thought of a charge pump (sorry about the water analogy).
Initially the plates terminals are uncharged and the charge pump moves charges from one terminal to the other terminal making one terminal having a net positive charge and the other terminal having a net negative charge.
The net charges on the terminals produce an electric field within the cell which opposes the pumping of charges from one terminal to the other by the charge pump (electrochemical process).
Eventually that electric field is enough to stop the migration of charges between the two terminals.
Each cell has a certain potential difference across its terminals and the addition of these potential differences is the same as for the example with capacitors.  
So what is different about the cells as compared with the capacitors.
The key difference is that as soon as the terminals of a cell are connected together with a conductor charges will flow from one terminal to the other in the external circuit.
The reduction in charge on the terminals would result in a reduction in the electric field between the terminals but that electric field is no longer sufficiently strong to stop the charge pump pumping charges causing a migration of charges between the two plates within the cell.
The charge pump moves charges within the cell to compensate for those charges which flowed from one terminal to the other terminal via the external conducting circuit.
In effect the charge pump maintains the potential difference across the terminals whereas with a capacitor there is no mechanism for maintaining the amount of charge stored on the plates.  
The removal of charges from terminals $D$ and $A$ would result in a migration of charges between terminals $C$ and $B$ ie an electric current flows between the two cells.
A: Discharge cycle:
Because of electrochemical processes, a battery develops an electric potential difference (voltage) between its electrodes.  When you connect two such batteries in series, each battery maintains its individual potential difference (its voltage).  When the circuit is completed, by connecting a load between the positive electrode of the first battery and the negative electrode of the second battery, a current, determined by the sum of the voltage of the two batteries (2x) divided by the load resistance added to the internal resistance of the batteries (Rl + 2Ri), flows.  This process continues until the batteries are discharged.  
If we designate the negative terminal of the second battery as the ground state, then the connection between the two batteries will be at (x) volts (with respect to ground), and the positive electrode of the first battery will be at (2x) volts (with respect to ground).  
Charge cycle:
Since the internal resistance of the batteries is the same, the voltage (2x) provided by the charger is divided in half.  This is equivalent to each battery receiving x voltage. The first battery will charge to x voltage and so will the second.  When both are charged, the current will be 0, since the voltage of the batteries will equal the voltage of the charger.   
