Current in the legs of a MOSFET

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.}

The body terminal is there whether we like it or not because of the construction of a MOSFET:

The gate voltage modifies the channel width. But the gate potential is measured relative to the substrate potential, and the channel is formed in the substrate material. Therefore the MOSFET behavior depends strongly on the substrate (body) potential.

If we did not make a contact to the body, it would be free to float, and we could not control the behavior of the transistor. We connect to the body to take control of this behavior. Normally we tie the body to either a most-positive or most-negative potential, like you say, because that gives the lowest |V_gs| relative to the appropriate supply and prevents forward biasing the source-body junction. If we did not contact the body, it would still be there, but we would not have control over its effect.

(MOSFET image by Wikimedia user CyrilB licensed under CC-BY-SA)