OK, so if you directly connect the two terminals of a battery, there is very little resistance in the circuit and you will drain the battery very quickly. However if you have a very simple circuit, for example a light bulb connected to the battery, the battery will drain slowly based on the watts used by the light bulb. My question is, if this light bulb provides only a little resistance, wouldn't a lot of current flow through and therefore drain the battery quickly?
You may be a little confused by the resistance of a light bulb. Typically, a "cold" (that is, not glowing) bulb filament will have a resistance about 1/10 its resistance when hot. So an operating (glowing) bulb will discharge a battery much less quickly than you might think if you just go by what a meter says the resistance is.
Once it's glowing, another effect comes in to play. For a given desired brightness, you can get better efficiency by overdriving the bulb. That is, if you take a bulb and drive it at higher voltage, it gets even hotter than it would normally, and its resistance gets even greater. Since it is hotter, a greater fraction of the energy being dissipated by the filament is given off as visible light as opposed to infrared. See "blackbody radiation" for an explanation. Of course, this means the filament will burn out more quickly, but you can't have everything.