Electronic – MOSFET symbol with no arrow but circle instead

This is a somewhat artificial problem, because you've shown no mechanism by which the capacitor is initially discharged so as to have no voltage across it, so it makes it difficult to intuitively understand the problem. But I'll imagine there is also an ideal switch across the capacitor that is closed for t < 0 and opens at t = 0.

After t = 0, the supply V_S plays no further role in the behavior of the circuit on the other side of the capacitor, and we can ignore it.

When the FET turns on (I'll assume you mean that V_G is high enough to put the MOSFET in saturation), you can treat its channel as a very low value resistor.

That means you will have a very low voltage at the drain, given by the resistor divider equation:

\$ V_D = V_1 \frac{R_{FET}}{R + R_{FET}}\$

Where I've arbitrarily named the node at the FET's drain as "D" and the node between the capacitor and the resitor as "1".

V₁ will decay exponentially toward ground as usual for an RC circuit, and V_D will decay proportionally.

No, don't use thermal relief vias. The reason they are that large is so that they WILL fill up with solder, that will make a strong heat connection to the other side of the PCB. I have done several PCB's using this method, and it works well, the MOSFETS will solder just fine.

The PDF used 1 oz for its simulation, which usually is 1/2oz base and 1/2 oz plate. I would recommend using at least 2 oz copper, (1oz base, 1oz plate) if you have other smaller SMT parts on the board, or a heavier plating if just the MOSFETs. If you can get 3oz or 4oz, you will have a much better heatsink.

Look on page 17 of that PDF, more holes used to link the layers the cooler the MOSFETS ran because they had better thermal connection to the other layers of PCB.