Gets a little tricky. Let's say for simplicity that your coil is 4Ω. To halve the voltage the battery puts out, you could put a 4Ω resistor in series with the coil and each would drop half the voltage. Each cell is 4.2 volts fully charged >= 8v for both. 8v / 8Ω = 1 amp (roughly). Half the 8v across the resistor (4v) * 1 amp = 4 watts. The duty cycle of vaping is pretty low (5 seconds per minute), so you wouldn't need a resistor capable of continuous 4 watt power handling, but you would need a one watter or two.
However, you would be buying and charging two cells for the exact same performance as one (which would also be without the extra heat wasted by the dropping resistor). Paradiso mentioned voltage regulation and that is the fancy solution. One problem, though, is that the common regulators drop about 2.5v, so a 5v regulator would start dropping the output voltage when the cells output fell to 7.5v. The ultimate solution is to get a low dropout regulator (only a .5v drop between input and output). Here's the datasheet-
http://www.national.com/ds/LM/LM2940.pdf
and here's the part at digikey for $2.15-
Digi-Key - LM2940IMP-5.0CT-ND (National Semiconductor - LM2940IMP-5.0/NOPB)
Stick with the sot-223 chip at 7mm square versus the TO-220.
Now that you have a rock solid 5 volts, you have something that almost no one else has- steady vape production throughout the battery charge. Now, to knock down that last volt to get to 4v, just put a 1Ω 1/2 watt resistor in series with the coil. Voila, a solid 4v power source.
However, you also have no way to tell if your cells are over-discharged, and that will ruin your cells and create a danger. Bummer, now we need an undervoltage cutoff circuit.
Or, you could just wrap tin-foil around a dowel, cover that with plastic tape and go with one cell. Grin.
