Electronic – Darlington biasing design error in a commercial frequency counter oven control circuit

In short: you can't. The 0.6V threshold for a BJT is a consequence of the physics of silicon P-N junctions.

A germanium transistor would work, but you will have to mail-order it, and it will be expensive.

A rail-to-rail op-amp indeed may be an option.

However, another solution is to make the voltage of your audio signal higher, rather than making the transistor threshold lower. You could do this two ways:

Make the emitter voltage lower

^{simulate this circuit – Schematic created using CircuitLab}

Now, the audio signal is 0.6V higher than the emitter. Of course, you'd have to come up with a way to get a 0.6V power supply, and probably adjust it to get just the action you want. There's another way...

Add a DC bias to the signal

^{simulate this circuit}

Here you can adjust the pot to add some amount of DC bias to the signal to get the sensitivity you desire. The capacitor serves to isolate this DC from your audio source while allowing the AC signal to pass. This is called capacitive coupling.

R4 exists to limit the base current in case R1 is adjusted too far. There's no point in biasing the signal above 0.7V since that would mean the transistor is always on, so R4 also makes the useful adjustment range of R1 wider.

Also, notice in both cases I've added a resistor to the transistor base. You don't want to make this mistake.

I see two problems here:

1: Tweeters can be rated for total system power, not their own individual power. Pay very close attention to that when you read the specs. Given a full-range audio signal, most of the power (80-90%, more for bass-heavy stuff) goes to the woofer(s) and the tweeter only has to handle what's left. So a 200W system-rated tweeter may only be good for 10-30W on its own.

2: Per your actual question, that circuit won't gain anything from the higher supply because it's set up as an emitter-follower. For that design, the only benefit of a wider supply range is to prevent it from clipping because of the supply. It will never exceed the voltage that it's driven with.

That being said, you can easily level-shift a square wave using an NPN to ground and a pull-up resistor to the higher supply voltage. The low-level signal generator only sees the NPN and doesn't know about the higher supply, and the pull-up resistor only has to be strong enough to drive the base of the power transistor. Given enough current gain in the power stage (I see you have darlingtons, which are known for exactly that), this shouldn't be hard either.

This will also invert your signal. Inversions often come for free and sometimes have to be explicitly reversed. If you don't care about that, then you're done, else you can add a second transistor + resistor to invert it back. Only the last stage needs to share the higher supply.