Hi Mr. P,
I'm no pro but I'm gonna take a shot at going into some detail about what's going on here. As much for my own education as yours. If I blow something here I hope to be corrected.
The subcircuit in the top right corner is creating a virtual ground from a single voltage supply. The idea is that many opamps require a split supply (i.e. instead of just wanting a single supply, say, +15V and ground, they want +7.5V, ground, and -7.5V) but you often just have a single supply available. One solution is what they've done here: take the positive voltage, create a voltage halfway between it and 0V, and call this new middle voltage "ground". R12 and R13 create what's called a "voltage divider", and since they are of equal values, the voltage at the junction between them is halfway between Vcc and -Vcc.
I think it's accurate to say that this works because voltages are relative, not absolute: i.e. when we say a certain point in a circuit is at +5V, what we really mean is that it's 5V more positive than some other point in the circuit (if not specified, usually ground).
If you look at the main circuit, you'll see that the op-amp has connections to Vcc and -Vcc, which means it connects at those points to the points marked Vcc and -Vcc in the top-right subcircuit--which are the battery terminals. The middle supply from the subcircuit is then called "ground" or 0V and is connected to all the points in the main circuit marked ground (inverted triangle thing).
Here's where my knowledge gets a bit shakier but I'll take a stab at it. The main circuit is a simple active mixer (adder) with gain of 2.2 (R10/R1). R1 and R2 present the inputs with an impedance of about 10k. I don't have a lot of experience with these things but that seems low to me. R1 and R2 then join up to the op-amp's inverting input (marked '-'). Gain is set by the ratio of R10 to the input resistor (either of R1 or R2 in this case).
Capacitor C10 sets the high-frequency cutoff for the circuit. This allows lower frequencies to be amplified while the higher frequencies we don't want are attenuated. This is also called a "low-pass filter", since it passes lower frequencies more readily than higher ones. The filter is described by its "cutoff point" or "break-point", which is usually taken as the frequency at which the gain is -3dB. The formula to find the break-point is f = 1/(2piRC). In this case, R is R10 and C is C10, so the break-point is at 1 / (2 * pi * 22,000 * (47*10^-12)), which if I'm doing this right puts the break-point at 154 kHz.
If I'm doing that wrong will someone please beat me with a clue stick?
I'm not sure what R11 is for. C11 blocks DC from going to the output (it "capacitively couples" the output).
I hope this helps! I also hope I got most of it right. . .
Torben