Problem with driving the mosfet through optocoupler

The way it's hooked up now it looks like at best the gate of the FET will be at half the drain voltage. If there's enough voltage on the drain it might get into a linear region where there's some drain current flowing, but it's not going to fully enhance the FET. You really want to hook the side of R2 that's attached to the drain to an independent supply voltage that can give you enough Vgs to turn the FET on fully. (For example to the + side of the 9V battery, not the drain of the FET.)

The reason the FET turns off so slowly is because you only have 2 kΩ pulling it high. Take a look at the datasheet for that FET. It should show you the actual and effective gate capacitance when switching. The relatively weak 2 kΩ pullup is working against that capacitance.

Here is a trick I sometimes use in this situation:

The double emitter follower is basically a impedance buffer. The current to charge and discharge the gate capacitance is handled by Q1 and Q2, so the signal at GATE can be much higher impedance and still switch the FET quickly. Note that this double emitter follower will loose 700 mV or so on each end. That is fine for a FET that switches over a 10 V gate range. It will still be "off" with 700 mV on the gate just as well as with 0. At the other end, its usually easy enough to drive GATE the extra 700 mV further than what you want to make sure shows up on the gate of Q3.