Electronic – Potentiometer for two 12V/1.1A fans in parallel

1. No. A dead spot means there is effectively a large resistance in series with the wiper, not that it shorts to anything.
2. C1 is to filter out some of the high frequencies that are clearly not part of a valid signal. Even in a hurricane, there is some rather low (relative to what even simple and cheap electronics can handle) maximum signal frequency on the pickup coil. C1 probably just shorts out things like radio station pickup.
3. D6 and D7 are clamps. They should never conduct in normal operation. They are there to protect things in case of a screwup or failure elsewhere that could put too much voltage on the pot.
4. It seems from your description it is only on one side. Measuring it would tell you for sure, of course.
5. Your quote is out of context, so this is hard to answer. In general, expect spikes and other noise to be present on external signals that come in from off-board. This seems to be just telling you to use normal precautions for external signals, like make sure your device doesn't blow up if one of the signals is accidentally connected to the power line, or gets zapped by a static charge.

Each of the 3 power rails (3.3V, 5V, +12V) are connected to a fuse, then to a binding post, while the negative binding posts go to ground.

This sounds like the problem. Keep in mind, voltages are differences in potential between two points. There's nothing special about ground, it's just an arbitrary point which we pick. It's 0V because the difference of something with itself is 0.

You can call anything you want "ground". That's why this circuit works. If you call the -12V output as "ground", then everything else is 12V higher. That's including what was previously called "ground": now it's 12V, because it's 12V more than what you are now calling ground.

Now consider what you've done:

^{simulate this circuit – Schematic created using CircuitLab}

The "ground" of the power supply connects all of the voltages supplied together (connections labeled B and C). The output voltages are relative to this. Notice how the -12V (V4) makes a negative voltage because it's positive side is attached to "ground".

Then, you attached the negative binding posts of all the supplies together. Largely this is redundant: you are duplicating connections B and C. But you are also adding connection A.

See the problem? You've shorted out V4. A wire has ideally zero resistance. By Ohm's law, the current that will flow is:

$$ \frac{12V}{0\Omega} = $$

In reality, the wires used to make this connection actually have some very small resistance, and you get a whole ton of current. This far exceeded the current the voltage regulator can handle and the smoke got out.