Electronic – Why place a diode in series with the LED in an optocoupler

You can't use Ohm's law to calculate the current through a diode.

However, the resistor is in series with the diode and, as you've learned, series connected circuit elements have identical currents.

Thus, you can use Ohm's Law to find the current through the resistor and diode but, to do that, you need the voltage across the resistor.

When you subtract the diode voltage from the source voltage, by KVL, what remains is the voltage across the resistor \$V_R\$.

Now, if you know the desired current, e.g., \$20mA\$, you can calculate the resistance required for that series current.

Thus, by Ohm's Law, the desired resistor value is:

$$R = \dfrac{V_R}{20mA} = \dfrac{V_S - V_D}{20mA} = \dfrac{9V - 1.7V}{20mA}$$

The diode is there to discharge the capacitor quickly.

If the output of U3A is high the diode is reversed polarized, so you can ignore it then. Then C7 gets charged via R22 with a time constant of 4 ms. That means it takes about 20 ms to charge completely.

If the output of U3A is low the diode conducts and discharges C7 very quickly, which you can see as the sharp falling edge on the scope.