When you open a faucet, you will get to the point where as much water is flowing through the faucet as the inlet water pressure will allow. That's saturation. (Think of the inlet connection to the faucet as the collector, the part you turn with your hand as the base, and the part of the faucet that lets water flow into the sink as the emitter. There is actually a water valve that more exactly resembles a transistor, called "eductor.")

A transistor is a valve. The base current (obviously related to base voltage, but bipolar transistors are more easily described a current valves) controls how much current flows from collector to emitter. Under normal amplifier operation, emitter current will be the sum of base current and collector current, but current gain "beta" makes it so that base current will be small fraction of the total current flowing out the emitter. Most of that emitter current comes from the collector. In this normal operating range, the voltage at the base connection is lower than the voltage at the collector. The base to collector junction is reverse biased. The base to emitter junction is forward biased.

When in saturation, maximum current flowing between collector and emitter, voltage Vce is at its lowest (obviously, otherwise maximum current wouldn't be flowing!), the voltage at that collector is now actually lower than the voltage at the base, so the base to collector junction is forward biased. The base to emitter junction is too, but that's essentially always the case, so nothing unusual there!

Take a look at the second explanation here. I think it's very well done.

http://electronics.stackexchange.com/questions/13063/how-do-i-saturate-an-npn-transistor

This is another good explanation, although not any clearer than the Stackexchange one (IMO).

http://ecetutorials.com/analog-electronics/operation-of-bjt/

Or are you looking for a more semiconductor physics type of explanation?