Do I need a buck converter?
The Arduino has a linear regulator, which means the current in = current out, and the excess voltage translates into wasted heat in the regulator.

Your datasheet shows that the receiver module will use a maximum current of 20mA. Let's assume it uses this all the time.

If you use a buck converter, it will use something like 10mA from your power source (a battery?). 12V * 10mA = 120mW. 5V * 20mA = 100mW. The difference is 20mW which translates to heat in the buck converter. 10mA is just an example number that I have made up - if you have a particular regulator then the datasheet should allow you to estimate how much the input current will be.

If you use a linear regulator, it will use 20mA from the power source. 12V * 20mA = 240mW. 5V * 2mA = 100mW. The regulator will heat up by 140mW.

This question tells me you can get 900mA through that pin when using a 7V power supply, which is 1800mW power dissipation (900mA * (7V - 5V)). So 140mW should be no problem for the regulator.

You might want to use a buck converter if:


You care about wasted power (more power = less battery lifetime!)
You care about heat (maybe if this circuit is designed to work in the desert at noon, or if you want to put it in a tiny box with no airflow)
You want to experiment with buck converters


Otherwise, I don't see any reason not to use the Arduino's built-in linear regulator.