Understanding how an audio amplifying circuit works
Does this circuit look correct? (it physically works, but I don't know
  if I drew the circuit correctly :/ )


Yup. Looks reasonable. At least, as reasonable as the circuit is, which isn't very.


  What does the capacitor actually do in this circuit?


It allows the FET to ignore the DC level on the input. So, you can present it with 1 VAC around ground, or 1 VAC riding on 5 volts, or whatever. As I'll get to in a moment, this can be important.


  Why do I need a resistor across the gate and drain legs?


Now, this is where it gets interesting, and not entirely in a good way.

The FET works like this: above some threshold (called Vgs(th) on the data sheet), increasing the GS (gate-source) voltage, causes the apparent resistance between DS (drain-source) to drop. In this circuit, this causes the current in the speaker and the FET to increase, which causes the drain (and hence the gate) voltage to drop. At some point, the two effects will stabilize, and this arrangement is called negative feedback.

Here's where this circuit shows its problems. Remember the threshold I mentioned? For your part it's in the range of 2 to 4 volts. Let's say 4. So, if the gate voltage is less than 4 volts the FET will be off and a full 9 volts available at the gate. This will cause the resistance of the FET to drop, and at some point (about 4 volts) the FET will be turned on enough to give 4 volts at the gate. Problem is, this means that the speaker is dropping 5 volts. With an 8 ohm speaker, this means that, with no input, the circuit will draw 5 volts/8 ohms, or about 0.6 amps. This will quickly drain any battery you use for power. Not only that, it will produce about 3 watts of heating in the speaker. Since your maximum possible music output is 0.7 (2.5 ^2/8), or about 0.54 watts, this is a grossly inefficient system.

Now, as to R2. It more or less isolates the AC part of the gate signal from the drain. Without it, the signal passing through the capacitor would effectively have to drive 8 ohms, and this would produce a very small signal at the gate, since you would have a voltage divider of 8 ohms/10,000. 

ETA - Oh yes, and I forgot. Remember when I mentioned the utility of a blocking capacitor (C1)? While this circuit is not a good choice for a power amplifier, something like it is quite useful as a preamplifier to provide an input for a power amplifier which IS a good choice. Such a preamplifier circuit will (unless it also incorporates a blocking capacitor on the output), have a large DC level on its output. About 4 volts in this case. Or it may be centered around zero if a more sophisticated circuit is used. Since you never know what sort of weird circuit may be hooked up to an amplifier, putting a DC blocking capacitor on the input is a very good idea.