Decoupling Capacitors: What happens to the DC on the cap once noise is introduced and it shorts?

The capacitor does not "short out", it has charged up to a constant voltage by storing energy as electrical charge, and if something external tries to change the voltage over the capacitor, it means that more or less charge is needed to change the capacitor voltage up or down, and moving charges means is current flowing.

So in short, a capacitor wants to keep the voltage over it constant, and resists any voltage changes by combating it with current. So voltage spikes get attenuated because the capacitor uses energy of the spike to change charge, and the larger the capacitance is, the less the spike can change the capacitor voltage.

The fundamental nature of a capacitor is that stored charge = capacitance x voltage: - $$Q = CV$$

We also know that current is the rate of change of charge with time hence, if the formula is differentiated with respect to time we get this: -

$$I = C\cdot\dfrac{dV}{dt}$$

The impact of this formula is that if a current is injected, the capacitor's terminal voltage will ramp up or down depending on whether the current is positive or negative.

If that current were applied to a resistor, there would be a step change in voltage but, for a capacitor, there is a ramping up or down and not a sudden loss of terminal voltage.

Looking at it from a different angle, if a voltage ramp is applied to the capacitor, there will be a constant current flow into and from the capacitor.

Neither of these scenarios imply that the capacitor turns into a short circuit other than to try and counteract changes to its terminal voltage; there can be no sudden fall to zero volts because that would imply an infinite current surge into the capacitor.

Imagine the capacitor like a flywheel rotating at constant speed; any action that attempts to speed-up or slow-down the flywheel requires great torque and that results in only a ramping up or down of speed and not a sudden change of speed to zero.

The capacitor does not become an actual short circuit.

When we say that a capacitor is a short circuit at high frequencies, we are talking about the impedance of a capacitor when a sine wave voltage is applied to it. The impedance is the voltage divided by the current (similar to resistance).

$$ Z_C = \frac{V}I = \frac{1}{2\pi jfC} $$

It's an imaginary number, which means the current and voltage are out of phase. Apart from that, impedance is very similar to resistance.

Let's say you have a 1uF capacitor and a 1V (rms), 10kHz sine wave. The math says the impedance is 15.9 imaginary ohms. Therefore the current is \$\frac{1V}{15.9\Omega} = 0.063A\$. Now calculate the current at 100kHz. The current is \$0.63A\$. Now try 10MHz. The current is \$63A\$. Lots of current!

A real short circuit would allow infinity amps to flow, of course. As we increase the frequency, the impedance gets closer to 0 and the current gets closer to infinity. That's what "capacitors are like short circuits at high frequencies" means. A high frequency voltage signal will cause a large current to flow. (Or, equivalently, a high frequency current signal will only create a small voltage)

So what happens if you have a signal consisting of several frequencies? Capacitors are called linear elements, which means that we can analyze each frequency independently. If you have a DC voltage (0 Hz), and a high frequency noise voltage, then the DC signal will not create any current through the capacitor (the impedance is infinite), but the high frequency noise voltage will create a large high frequency noise current through the capacitor which acts to cancel out the noise voltage.