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We know that the normal electric field direction in a parallel plate capacitor normally goes from the positive plate to the negative plate, so going from a high potential to low potential. Thus, if we were to place a positive charged object into the plates, the positive object would go from a higher potential to lower potential, as the object would start with potential energy and end with kinetic energy. However, if we place a electron in the capacitor, the electron has a high potential energy in the negative plate... and accelerate towards the positive plate. Would we say that it is going from a lower potential to a higher potential in terms of the parallel plate, or would the field follow the energy flow of the electron, going from high potential to low potential?

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Since potential does not depend on the "test charge", then yes, we say that the negative charge is going from a lower potential to a higher potential. i.e. the negative charge just moving in the field will experience a positive potential change $\Delta V>0$

However, since the change in electric potential energy is given by $\Delta U=q\Delta V$, if $q<0$ and $\Delta V>0$ then $\Delta U<0$. So the negative charge still loses potential energy, which is needed since it is gaining kinetic energy, and the electrostatic interaction is conservative.

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No, you can't say that the electron increases it's potential energy. It may be true regarding the parallel plate capacitor to say that the electron moves from a region of lower potential to a region of higher one (considering the convention).

Any charged particle will tend to move from a region of higher potential energy to a region of lower potential energy.

It's just a matter of convention that the electric field lines are drawn in accordance with the behavior of a unit positive test charge. If you want your test charge to be an electron, just reverse the direction of the electric field lines.

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Potential energy is given by qV. Q for an electron is negative and by moving from the negative to the positive plate, it will have negative potential energy gain or will lose potential energy.

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