Current loops were initially designed to connect multiple teletypewriters in series (originally 60 ma was used instead of 20 ma). This scheme was latter adapted for instrumentation use, with 4-20 ma as the standard.
In your current receiver design, one end of the resistor is connected to ground. Instead it should be passed on to the next receiver, like the old series Christmas tree light strings. The last receiver connects to a lead which goes back to the transmitter ground.
At each receiver, a 250 ohm resistor is placed in the loop. The two sides of the resistor can then be fed into an optical coupler, which will turn a transistor on/off which you can feed to your circuit. (Or if you need to still be able to power your sensor from the loop, you will need to make the sensor run off a floating ground reference tied to one end of the sensor.)
If the sensor requires a minimum of 12v, using a 250 ohm resistor this implies a current of 48 ma, not 20 -- so I'm not sure if this is really a 4-20 ma loop or not.
If you have to use the current receiver design, it should be placed last. All of the others must allow the loop to be passed on. (Which limits you to two receivers, one new and one original design if you are allowed only one new receiver.)
Best Answer
You can just use a 150 ohm 1% resistor to ground and leave out the rest.
The Expressif chip has about a 1V input range, but the boards typically have divider resistors so the range is actually about 3V (maybe nominally 3.3, but you can't count on that due to reference tolerance and resistor tolerance) because of an internal divider.
The ADC has appallingly-bad characteristics (linearity, stability, noise, tolerance) and is only worthwhile because of the price (which is zero). You should research this before going any further down this path.
Tying it to any decent 4~20mA transmitter is probably a gross mismatch, suggest you look at using an external ADC.