Voltage regulator VS power supply

A switching power supply will be noisier than a linear regulator, no question. The only way to know if it will disturb the wifi is to power it up and see what happens.

Any switching regulator will dissipate power as a function of conduction losses and switching losses. These parts with an integrated MOSFET do need PCB cooling - usually a multilayer PCB with relatively large 'islands' of copper interconnected with vias. Again, you'll have to do some math to figure out a loss estimate, and gauge the copper size accordingly.

The datasheet says that the 5V part at 2A load and 12V in operates at around 83% efficiency. So, for 10W out you're losing just over 2W in the device (and the external diode that completes the buck converter). This will scale down somewhat with your reduced loading (conduction losses will drop, switching losses probably won't).

You are wasting quite a bit of power by using linear regulators to drop 9 V to 5 V and particularly to 3.3 V. There is also no need to run the LED from 9 V. There is no such thing as a "9 V LED". There are some packaged devices that contain a LED and resistor so that the whole thing is intended to run from some particular volage, like 9 V, 12 V, 24 V, or whatever, but then you don't have just a LED anymore.

It would be useful to know the current requirements of the 5 V circuitry, the 3.3 V circuitry, and the LED. I'm guessing that the sensor probably doesn't take much current. In that case, I'd probably use a linear regulator to make 5 V as you suggest, but a switcher to make the 3.3 V from the 9 V. The sensor runs from 5 V and everything else, including the LED, runs from 3.3 V. That will be a lot more efficient. Assuming this is a normal red, yellow, or green LED, it can be run from 3.3 V easily with the right resistor in series.