Why did my LED resistor burn while lighting four LEDs in series?

6.8 volts seems awfully high for a single LED. Are you sure that 6.8 is not the number for all four LEDs? That would make it 1.7 volts per LED, which is more reasonable for a red LED. And that would mean that you are currently pushing 172 milliamps, or almost 3 watts through your resistor.

If that is the case, you should lower your power supply to less than 20 volts (maybe 12 volts) to keep from destroying the gate of your MosFET (M2).

I see your problem. Your circuit shows how you're driving a single LED segment. (I presume you then have 7 of these circuits, one for each segment.) The datasheet shows 4 LEDs in series, covering the segment.

Where you've gone wrong is assuming there's 6.8V forward voltage drop per LED. There is no such red LED. Typically a red LED will be around 1.6V-1.8V forward voltage drop, and that's a characteristic of the physics involved so there isn't really much scope for variation. This tells me that you have 6.8V forward voltage drop for all four LEDs in that segment in series.

So with a 6.8V voltage drop and a 24V supply, you're dropping 17.2V across the 100R. As Mark says, this gives you 172mA and 2.96W power dissipation on the resistor. Not healthy for a 0.25W resistor.

In fact you're lucky that the 0.25W resistor basically becomes a fuse under those conditions and burns out almost immediately. If it hadn't, putting 172mA through the display would burn that out pretty quickly, and a large 7-segment display is going to be a fair bit more expensive than a resistor. If you'd used a higher-powered resistor, you'd be wondering why the display briefly flashed very brightly indeed and then went black forever.