The answers of Martin and Edward
are quite right, but theý lack a componet which plays a role
in the case of that "garage opener battery".
If current is not limited in the circuit outside the battery,
first internal resistance can be the limiting factor.
But, often in such small (the measure for "small" is the area/volume of
active material at the battery poles) batteries
the current is limited by the speed of chemical reaction at the
electrodes. This is not a linear function of current, thus
it cannot viewed as a part of internal resistance.
Does it have anything to do with
contact area? If it does, then how
would you model it?
Yes, contact area is important, but this is a problem at high
current densities mostly.
Depending on voltage in such a circuit, small contact area
(= high current density)
can result in contact welding, sparking, or ignition of an arc,
or in some strange semiconductor effects of the contact point.
(metal surfaces are nearly always coated by some oxide)
Did You ever watch when an electric arc is started by a welder?
This is main business when designing contacts in relays, swiches in
home, or the switches in 230 kV lines.
If you take the simplest form of capacitor, two parallel plates, the the capacitance is proportional to the area of the plates and inversely proportional to the distance between the plates:
$$C \propto \frac{A}{d}$$
When you're making a capacitor out of a snapple bottle you're actually making something similar to the simple "two plate" capacitor but the "plates" are curved round the surface of the bottle, with the foil as the external plate and the (conducting) salt water in the bottle acting as the internal plate. The $d$ in the equation above is the thickness of the glass.
So you can make a capacitor out of any bottle, jar or anything similar. All that matters is the area of the foil and the thickness of the glass. If you want to increase the total capacitance you can just link any number of bottles in parallel i.e. link the external foil covers as one electrode and the internal brine solution as the other electrode. When you join up capacitors in this way, "in parallel", you get the total capacitance just by adding up the individual capacitances of all the bottles you've joined.
You ask about the effect of the increased voltage: the charge stored in a capacitor is given by:
$$Q = CV$$
where $C$ is the capacitance and $V$ the voltage, so using 12kV instead of 9kV just means you get 33% more charge for a given capacitance, or alternatively you can get away with a smaller capacitance to hold the same charge.
Do I sound a bit like a grandma if I point out you need to be careful with this experiment. A typical Tesla coil can't generate enough current to kill you, but if you gang together enough capacitors the stored charge in them will kill you!
Finally, I normally point people to Wikipedia if they want to learn more, so see http://en.wikipedia.org/wiki/Capacitor and http://en.wikipedia.org/wiki/Leyden_jar for the sort of capacitor you're making. However be warned that the Wikipedia article on the capacitor is a bit technical.
Best Answer
Your question is partly physical chemistry (electrochemistry) and partly biochemistry, and both suggest the question is off-topic and should be moved to either the Chemistry SE or some Biology or Medicine site (I think the Biology SE got closed).
But since you ask:
When you contact together different metals you will generally find that electrons will move from one metal to the other so one metal will be oxidised and the other metal will be reduced. Because we have electron flow there must be a potential difference between the metals, and this potential difference is called the electrode potential.
The situation in the mouth will be complicated because the aluminium foil will be covered in aluminium oxide, and the mercury in the filling is present as an amalgum (typically with silver) and will probably be coated with mercury compounds. A plausible reaction would be:
$$ 2Al + 3HgO + 3H_2O \rightarrow 2Al(OH)_3 + 3Hg $$
If this is the right reaction then from the list of standard electrode potentials the electrode potential for this reaction would be about 2.2V or more than an AA cell. However Google has failed to find me any actual measurements so you should regard this as a guideline at best.
Even though the voltage is quite high the actual currents are likely to be extremely small since the reaction will be very slow. At a guess, and it is only a guess, I'd be very surprised if the current generated could do any harm. You need to consult a biologist for a more informed answer.
Whether it is relevant or not I don't know, but I note that aluminium and mercury enthusiastically form an amalgum with each other (though the reaction requires unoxidised aluminum and is slowed by the aluminium oxide layer). This may enahnce the electric contact between the aluminium foil and amalgum filling and maybe enhance the current and therefore the pain.