yes, that is a NAND gate, but all you need to know is to hook pin 2 to your uC and drive it low to turn on the circuit. To set the pin 1 resistor, take the 5v source minus the on LED voltage (1.5V) minus the output low voltage (assume .7V), divided by the current.
Note the design specs are listed at 10mA, so, designing for 10mA, (5-1.5-.7) / .01 = 2.8V / .01A = 280 ohm = 270 ohm or 300 ohm.
{2.8V / 270 ohm = 10.37mA or 2.8V / 300 ohm = 9.33mA Note that the uC output low and/or LED Vf (voltage, forward drop) can be a little higher or lower, as well as the power supply, so that is why we design with 5% resistors when it doesn't matter if we use 10mA, +/- 10% (9-11mA).
As the note in the schematic states, the resistor/capacitor values are only representative of a specific design, so depending on what your load is will depend on whether or not you even need a snubber. Basically, you calculate the RC value needed for the frequency response you are looking at. Then, since capacitors come in 10% values (or larger), you pick a capacitor, then base the resistor value on that. The values given are a good place to start, and I'm am not an expert in snubber design, but I can research it. Unless someone else wants to weigh in here???
note that you need an extra snubber (rc network) when driving it high vs low, that's why driving the low side is more popular.