Post 4 said that it didn't, so case closed on that part. But why?

Sensor output is some voltage with some amount of current available. An IR output is the sensor output with a resistor in series with it to limit the current that flows through an IR LED.

Let's say the voltage out of the sensor is 5 volts that switches on and off in the pattern of the IR command. Let's say that 100 mA is available. 100 mA is WAY more than is needed to run four LEDs, the usual amount that you expect to get with a sensor and a block.

E = IR. I = E/R. R = E/I. Okay. Let's see here.

That means that the sensor looks like a zero ohms voltage source with 5/.1 ohms, or 50 ohms, in series with it. Another way to say this is that its internal resistance is 50 ohms. Whenever you figure what current is available through different circuits, you have to count this resistance as though it's in the circuit somewhere, even though you can't see it or measure it directly.

If you put an LED in series with this, the LED will drop around 2.0 volts, so you'll have 3.0 volts through the 50 ohms of the internal resistance, and about 3.0/50 = or 60 mA, enough that you might blow an LED. This coincides with experience.

Xantech distribution blocks use 470 ohms in series with the sensor output to feed LEDs. Then you have 3.0/(50+470) = 5.7 mA, a reasonable current for an LED that is stuck on top of a component sensor.

But what about running the LED signal into an RP?

If the RP's input impedance is, say 500 ohms, then that 500 ohms is in series with the 50 ohms source resistance of the sensor. This is a voltage divider; 500 ohms of the resistance is inside the RP, so 500/550 of the voltage will appear at the RP, being 4.55 volts. I can see how an RP could be made to work with that.

But if the sensor voltage goes through a 470 ohm resistor to become proper to feed an LED, then that voltage is fed to the RP, the total resistance of the circuit will be

50 ohms internal sensor resistance
470 ohms resistance for LED output
500 ohms input resistance of RP
total of 1020 ohms.

500 ohms is inside the RP, so the voltage at the RP will be 500/1020 times that 5 volts, or 2.45 volts. This is half the intended voltage, so probably won't work.

That's the nuts and bolts of why you often can't get an input to work if it's not specifically designed to take a signal intended for an LED.