Electrical – Why do you use a buffer when the whole point of op amp is to amplify signal

If you intend to run the LED from the same supply as the logic gate, and that gate can source and sink symmetrically, then it doesn't matter. Even though a logic gate may be rated for the same source and sink current, it is still possible that the low side driver has a little lower resistance. The total Vdd and Vss currents the device is rated for may also not be symmetric, even if individual pins are.

In general, it is more likely that low side drivers have a little less resistance and that the total Vss current rating is a little higher. Also, using the low side driver works with open drain outputs. This is why people are used to wiring LEDs between the output and Vdd. This means you want a inversion in there for a logic high to cause the LED to light. Of course if this is all coming from a microcontroller, which it almost always is these days, then digital signal polarity is irrelevant since it can be handled in the firmware either way.

As long as you are not violating any spec in the datasheet, go ahead and wire up the LED whatever way is most convenient for you. You could let the final board routing decide.

Yes, of course you can use RC4558 op-amps for line-level audio. They have been used over the past 30 something years in lots of low end gear. They work, but the quality isn't very good: poor noise and distortion figures, poor driving ability of loads in the less than 10K range. Use something better like an LM4562 or the inexpensive and good NE5532.

It's not a great idea to leave unused devices unconnected. Connect the output to the inverting input, and convert the noninverting input to your V/2 voltage reference (or the ground in a proper dual-voltage circuit).

From your description of the circuit, you are omitting important biasing details. Since you are working with a single power supply between 0 and 12V. The center voltage is at 6V. That is to say, when there is no signal, you want the output of the op-amp to be approximately 6V. This means that the non-inverting input will have to be at 6V. The easiest way to do that is to bias that input to 6V with some resistors, and then couple the signal with a capacitor, so that the signal's AC swings are superimposed on the 6V bias.

Legend:

• R1, R2 and C1 create a "phantom ground" reference voltage. The point between R1 and R2 is at 6V DC. Capacitor C1 ensures that this point has a low impedance to ground for AC signals, and also helps to reduce power supply noise at this phantom ground.
• R3 conveys the 6V phantom ground to the + input. Because it is a 10K resistor, it establishes a 10K input impedance (common for line level audio). R3 also has another important role: it provides a path for an input bias current to flow into or out of the + input. Without this resistor, the only path is through the capacitor C2, which blocks direct current.
• C2 couples the input signal to the + input, superimposing it on the 6V DC. The 5uF value is chosen so that it cleanly passes 20 Hz and up when combined with the 10K input impedance. Try to use a film capacitor. If you raise the input impedance, you can lower the capacitance. E.g. for a 100K input impedance, you'd need only a 0.5 uF capacitor for the same frequency response.
• C3 couples the output signal to the next output stage. This output coupling cap can usually be omitted. If the next device needs one, it supplies it.
• R4 and R5 are DC return resistors. If the previous or next device is DC coupled and sources a small current, these resistors complete the circuit for that current. Because otherwise your input and output are open circuits with respect to DC, due to the coupling capacitors. These resistors can usually be omitted, since most devices take care of their own currents internally. You certainly do not need R4 if the C3 coupler is omitted, because the output stage of the op-amp is a low impedance voltage source (sink for current).
• The unused U1B is connected as described earlier. The phantom ground is conveyed to its + input, and it is in feedback.

Note: when constructing such a device, do not route the jack sleeve connections to a ground trace on the circuit board, even though the circuit suggests that. That creates a "pin 1 problem". The jacks should be grounded directly to the power supply (or if there is a conducting chassis, to the metal chassis, which is then grounded). If there is radio-frequency interference picked up by the cable shields, this approach prevents routing it to the circuit board where it will easily find its way where it isn't supposed to.