Electronic – Random activation of transistor base by capacitive coupling (ghost voltage)

I might be totally wrong here but my excuse is the question is very confusing. The circuit shown above is trying to pull the 2.52 V line down to ground through the 10R resistor and an NPN transistor. The original (pre-hacked circuit) you describe was it this? Forward and Reverse are mutually exclusive as are left and right. On the vehicles I've hacked into they usually have a spring loaded centre off action (joystick). Are you absolutely certain the pole of the switch went to ground (0v)?

You say when you touched R2 (140k) the remote worked sending a signal to the car. This sounds like you were acting as an input signal. By touching the input your body as an aerial picking up the local EM field, the 140K resistor being large enough to prevent the weak signal from being shorted out. That suggests to me that the control input needs to go HIGH rather than LOW and that the switch was actually connected to the positive rail. To test this hypothesis connect up the circuit below and let us know what happened.

Edit: Actually, I found out the answer I posted is already given in the question you rejected as "too complicated". I ask you to reconsider that judgement, as anything involving operational amplifiers (suggestions also found in that question) will be even more complicated.

While the commentors are right that your circuit is a really bad idea in practice, your thoughts that lead you to the that circuit are actually quite good. You notice correctly that the voltage you obtain at typical volumes is too low to turn on the transistor (you need around 0.6V to get a significant effect). Using a darlington transistor doesn't help, as you experienced, because you need even more voltage to turn on a darlington transistor, but you can get away with a lot lower current. You problem is that your voltage is too low, not that your supply can't deliver enough current, so forget about darlingtons.

Increasing the "baseline voltage" as you call it, is exactly the way to go. Electronics people often call this process "adjusting the DC bias" of a transistor amplifier. It can be done quite easily on your circuit without adding a dangerous low-impedance power source like a battery. Let's work on your circuit. First, add a current limiting resistor into the LED circuit and put the LED on the other side of the transistor. It makes no difference in your circuit as it is now, but it will make soon:

^{simulate this circuit – Schematic created using CircuitLab}

You can fix your problem, as you correctly identified, by providing DC bias (a baseline voltage) to the transistor base. But you should not force a DC voltage into the headphone jack, so you need a capacitor for DC decoupling. This capacitor should pass the AC music signal, but block the DC required to bias the transistor. With the circuit as drawn, you can get the required DC bias (around 0.45V) from the battery using a voltage divider like this:

^{simulate this circuit}