Power Supply Output Voltage Query
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. (a) Charging the C1. (b) Transferring charge to C2.

For simplicity I will ignore all diode drops in this explanation.


The 10 V secondaries of your transformer will peak to \$ +10 \sqrt{2}~V \$ and \$ -10 \sqrt{2}~V \$.
At peak voltage the top end of C1 will be at +14 V and the bottom will be pulled to -14 V via D3. The capacitor will be charged to 28 V.
As the polarity reverses the voltage at the top of C1 is pulled down. The voltage at the bottom will attempt to maintain the 28 V difference. D3 will be reverse biased and will effectively be out of circuit (Figure 1b).
As the top of C1 reaches -14 V the bottom will have been pushed down to -42 V.
C2 will charge to -42 V via D6.


Your circuit has two of these charge pumps running on alternate half-cycles to improve the output current.
Drop a few volts for the diodes and you've still got enough to run the LM337.