For an N-Channel MOSFET, current is switched from drain to source. But a P-Channel MOSFET works in the opposite way - in a P-channel MOSFET, current is switched from source to drain. See this appnote from IRF. Also, the anode of the integral body diode in a power MOSFET is connected to the source of an N-Channel, but the drain of a P-Channel. See this excerpt.
Basically, when you have a positive voltage connected to a load, and you want to switch it on and off, use an N-Channel MOSFET between the negative terminal and ground. Allow current to flow by applying a positive voltage that will saturate the transistor (10-12 for power MOSFETs, 3-5V for logic level). Turn it off by pulling the gate down to the source.
When you have a load with the negative terminal grounded (which is usually preferable; don't muck with ground if at all possible!), and want to apply or remove a positive voltage, use a P-channel MOSFET. Pull its gate up to the source (which is connected to V+) to turn it off, or pull it to ground (through an open collector output if your logic signal is less than V+) to turn it on (So that Vg is 0, and Vs is, say, 12V, therefore Vgs is -12V).
Depletion mode mosfets are less common, and usually only available in N-Channel. For N-Channel depletion mode, the gate must be pulled below the source (which is often ground). Stick with enhancement mode for most switching applications unless you need something strange.
This schematic shows both (enhancement-mode) configurations:
To identify the source and drain, look at the side which the arrow is connected to. This is the source. If you've got a physical component, a diode test on a meter is useful both for finding the switched current direction (Apply positive voltage to the terminal which the diode test identifies as negative) and for a basic test (not a guarantee) that the transistor isn't burned up. To differentiate between N-Channel and P-Channel, look at the symbol: N-Channel pointing iN.
The current can still flow through the "substrate" even though the channel is pinched. The reason why it saturates is that there will be a region of higher resistance of size proportional to the Drain-Source voltage, and therefore the resistance of this region will be proportional to the same voltage.
But as current is voltage/resistance, the dependence will cancel out and you'll get "constant" current.
From Wiki (emphasis mine):
Even though the conductive channel formed by gate-to-source voltage no longer connects source to drain during saturation mode, carriers are not blocked from flowing. Considering again an n-channel enhancement-mode device, a depletion region exists in the p-type body, surrounding the conductive channel and drain and source regions. The electrons which comprise the channel are free to move out of the channel through the depletion region if attracted to the drain by drain-to-source voltage. The depletion region is free of carriers and has a resistance similar to silicon. Any increase of the drain-to-source voltage will increase the distance from drain to the pinch-off point, increasing the resistance of the depletion region in proportion to the drain-to-source voltage applied. This proportional change causes the drain-to-source current to remain relatively fixed, independent of changes to the drain-to-source voltage, quite unlike its ohmic behavior in the linear mode of operation. Thus, in saturation mode, the FET behaves as a constant-current source rather than as a resistor, and can effectively be used as a voltage amplifier. In this case, the gate-to-source voltage determines the level of constant current through the channel.
Also, from the MOSFET operation description, under saturation:
Since the drain voltage is higher than the source voltage, the electrons spread out, and conduction is not through a narrow channel but through a broader, two- or three-dimensional current distribution extending away from the interface and deeper in the substrate. The onset of this region is also known as pinch-off to indicate the lack of channel region near the drain. Although the channel does not extend the full length of the device, the electric field between the drain and the channel is very high, and conduction continues.
Best Answer
A MOSFET always consumes power from the circuit. It has no mechanism to convert energy from some other form to electrical energy.
Therefore, the currents through a MOSFET always flow from a higher potential to a lower one.
This means, for an n-channel FET, if the drain is biased higher than the source, current will flow from drain to source (through the channel). If the source is biased higher than the drain, current will flow from source to drain (through the body diode).
If you turn on the FET, you may get parallel conduction through the channel and body diode, but both currents will flow from drain to source, because the drain is at a higher potential.