Use of schottky diode and tracking peaks of input signal with capacitors
What's the use of the Schottky diode 'D1'?



The first opamp amplifies the input signal by (-)100. Notice that the non-inverting input is held at half-supply and that the input signal is AC coupled by C3 and the output by C4. The output of the opamp should swing about the half-supply voltage.
The second opamp is different.


The non-inverting input is ground referenced.
If the input signal (from C4) goes negative the output will go positive and the gain is \$ - \frac {R7}{R6} = \frac {22k}{10k} = -2.2 \$.
If the input goes positive the output will try to go negative but D1 effectively reduces the gain to near zero. 

At this point we have a half-wave rectified version of the signal.




Figure 1. (a) The AC 'music' signal. (b) Half-wave rectified signal (dashed) and peak-hold (solid).


  The instructable says that the capacitor 'C5' is used to track the peaks of the input signal. What is the exact procedure of this tracking? 


Very simple. The capacitor holds the signal level high to give time to read it on the display. D2 prevents it discharging back through the opamp when the opamp output goes low. This is the same as a half-wave rectified DC power supply.


  Also, what would happen if this capacitor and the resistor R9 through which it discharges is eliminated from the circuit?



If both components are omitted the circuit would still work in a fashion but any transient peaks would be very brief and appear very faint on the LED bar meter. In a professional audio level meter this would be considered "a bad thing" as the engineer could miss peaks or transients that may exceed the maximum recording or broadcast level.
If R9 were put back in without C5 there wouldn't be any significant change.
C5 holds the peak but if R9 were omitted the capacitor would have no discharge path. The LED display would show the reading for the highest level in the audio so far and stay that way.
The combination of C5 and R9 are set to hold long enough to see but short enough to follow the signal reasonably well. In this case the time constant is \$ \tau = RC = 10k \cdot 10u = 100~ms \$ - slow enough for the eye to read but fast enough to track the beat of the music.


Rock on!