Question about using N-channel MOSFETs in a current sharing controller

No, also for P-channel MOSFETs it's charging them. The confusion probably stems from the voltage you see at the lower side, which goes to ground (or near it). But that voltage isn't important, a capacitor's charge is determined by the voltage across it:

\$ Q = \Delta V \cdot C \$

So decreasing the gate voltage increases the gate-source voltage difference, which increases the charge of the capacitor.

When you switch off T1, there's current flowing from +12 V through R2 to the gate to discharge it's capacitance.

edit re the update of your question dd. 2012-07-09(*)
Turning off means that you discharge the gate to +5 V, and this happens by current through R2 and D3. So you bypass R? but R2 is still the limiting factor. A solution would be to swap R2 and T1, so that there's more current/less resistance to discharge the gate than to charge it.

\$ \$

_{(*) I'm using the ISO 8601 standard date format here. We have user from all over the world and for some 9/07 means 9 July, for others it's 7 September. ISO 8601 is unambiguous.}

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).

Now, Lets say I have connected Drain to higher than Source and current is flowing through the body diode, But at the same time I want to turn on the MOSFET Source to Drain channel to allow current to flow from the channel from source to drain parallel with 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.