Driver circuit calculations
The circuit assumes that the FET driver is chosen such that it drives the MOSFET's gate so as to turn on the MOSFET "completely". In other words, it's source-drain voltage drop and resistance (RDson) will be so low relative to other things in the circuit as to be negligible. Therefore you can just treat it as a short-circuit for analysis in most cases.

Once the forward voltage drop across the diode (any diode) exceeds a certain amount, current will start to flow. Once this happens the voltage drop increases if current increases, but the voltage drop levels out and so does not change very much with current. So the simplest approximation is to assume a constant voltage drop (listed in the LED datasheet somewhere).

\$V_{LED}\$ is poorly named since it is not the voltage drop across the LED so I will henceforth call it \$V_{supply}\$ instead. I will then called \$V_{LED}\$ the voltage drop across the LED.

In your circuit, you will apply a \${Vsupply}\$ that is higher than \$V_{LED}\$ so the LED actually conducts current. Since we approximate that \$V_{LED}\$ is a constant when conducting, we don't need to consider about how the amount of current being conducted affects \$V_{LED}\$. We just assume it's the \$V_{LED}\$ in the datasheet.

So that means that the leftover voltage remaining in the circuit is:
\$  V_{supply} - V_{LED}\$

Where is this voltage made up? Obviously across the only other thing left in the circuit: the resistor (we assumed that the NMOSFET is fully on and negligible voltage drop and resistance so we just treat it as a perfect, ideal closed switch).

That means that \$ V_{resistor} =  V_{supply} - V_{LED}\$

What's the current in the resistor? It's:
\$ I_{resistor} = \frac{V_{resistor}}  {R} = \frac{(V_{supply} - V_{LED})} {R}\$

The resistor and LED are in series so that means \$ I_{LED} = I_{resistor} = \frac{(V_{supply} - V_{LED})}{R} \$ 

This is just a wordy way of doing Kirchoff's voltage law around the circuit.