To start with your last question: yes, it's a microcontroller (not a microchip, that's a confusing word since it's the name of a manufacturer of a.o. microcontrollers).
Unfortunately the supply current is not for you to choose. At 3.3 V it will be typically 0.35 mA, but it can be anything between 0.27 mA and 0.45 mA. And the 10 kΩ you calculated is the equivalent resistor of the receiver's power consumption.
The resistor in the schematic forms a low-pass filter with the capacitor, letting DC and low frequencies pass and filter out the higher frequency noise. That resistor will cause a small voltage drop from the power supply; the higher the current, the higher the voltage drop, due to Ohm's Law. So worst case the current will be 0.45 mA, and then the 1kΩ resistor will cause a 450 mV drop. If you supply 3.3 V then the receiver will have 2.85 V left. This is enough to operate, but you have to take into account that it will only output 2.85 V as a high level, and you have to check that this is enough for the microcontroller.
You're right about the working of a decoupling capacitor, and that's part of its job. Like I said the capacitor is also part of a filter. If you choose 1 kΩ for the resistor, and 100 nF for the capacitor it will have a cutoff frequency of 1600 Hz.
While I've never heard of this circuit before, I watched the video @user23711 linked. I think I can give you a decent explanation.
Essentially, you have an AC source driving your device with a resistor in series acting as a current limiter. The voltage measurement is simple. You are literally just measuring the voltage across the device under test, just like any other voltage measurement with an oscilloscope or multimeter.
As for current, you are measuring the voltage across the resistor and not the device under test (DUT) because voltage is (sort of) proportional to current in a resistor. Since the resistor is in series with the DUT, the current is the same as that of the DUT. However, the voltage across the resistor will be proportional to the current while the voltage of the DUT has an unknown relationship to current. That is, after all, the purpose of this circuit.
Now, if you had a current probe for your scope, you wouldn't need this test setup. Also, be mindful that the V/I curve for a resistor is not perfectly linear. If you are looking for high accuracy this might not be the way to go.
Finally, the reason for the AC source is to frequently enough cycle the voltage of the DUT. Then, when using XY mode on the scope, the IV curve of the DUT can be seen. But really, you could use any source you want (that won't break the DUT), and view the voltage and current curves with respect to time.
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