Electronic – How much is parasitic capacitance of soldering pads on PCB

Your equation:

\$ C = \frac{e_oe_rLW}{d} \$

Is the parallel plate capacitance equation and assumes that the plate size L*W is large enough and the gap size d is small enough that most of the e-field is captured between the plates. With your system, most of the capacitance will be in the field lines around the wires so the result will not be accurate.

Usually this would be simulated with software but as a first approximation you should use the capacitance of a two wire line. This will be some what inaccurate because teh conductors are a significant size wrt to the spacing. But it will be better.

\$ C = \frac{2 \pi e_r e_o L}{2 ln(2h/b)} \$

where h = 1/2 the distance between wires and b = the "radius" of the wires.

In your case the wires will not be circular (but they are not square either). but this is a good first approx..

Some open source software, that is kind of hard to use if you don't have the right tools is found by a search of "fast field solvers"

My only problem with the electrical example is regarding how the system may couple its output back into its input. Is this a simple case of simple feedback? How does simple feedback relate to capacitive coupling?

An 1 Mohm 0805 resistor will have self capacitance of about 0.2pF and this, when applied in the feedback loop of an op-amp will reduce the gain from the 3dB point of 795 kHz. This means that the transimpedance amplifier you are designing will not work well at 10 MHz because there is too much capacitance - and that's just from a single surface mount resistor - imagine what capacitance there is between tracks and if this is not properly catered for by good PCB design, you'll be lucky to get 100 kHz flat operation from the above circuit.

The above is the case of negative feedback causing a signal to become smaller on the output due to parasitic capacitance.

It seems that the output would need some ripple voltage to begin with that would be amplified.

The minutest noise will be enough to trigger oscillation if the feedback is positive. Noise is present in all components at temperatures above absolute zero.

Regards driving tracks having capacitance this is not directly related to the possible cause of oscillations and additionally the rule of thumb in your question doesn't take into account many, many factors such as the power capabilities of the driver, and the track dimensions and whether the track is terminated. At low frequencies there is no real rule at all.