Electrical – Can a capacitor charge and discharge at the same time

Why is the resistor voltage initially equal to supply voltage? Is it because there is no voltage going across the capacitor yet? Therefore, as there is no voltage drop across the capacitor, all the voltage from the battery is across the resistor?

Sum of voltages on the passive elements must add up to the supply voltage.

$$ V_{supply}(t) = V_{switch}(t) + V_{resistor}(t) + V_{capacitor}(t) $$

Because of the fact that \$V_{switch}(t) = 0 \$ and \$V_{capacitor}(0) = 0 \$, \$V_{resistor}(0)\$ must be equal to \$V_{supply}(0)\$.

2.What exactly does "the voltage developed as the capacitor charges" refer to?

When you apply a voltage difference between capacitor plates, one plate has more positive potential with respect to the other one. This initiates an electric field field between the plates, which is a vector field, whose direction is from the positive plate the negative one.

There is an insulating material (dielectric material) between these capacitor plates. This dielectric material has no free electrons, so no charge flows through it. But another phenomenon occurs. The negatively charged electrons of the dielectric material tend to the positive plate, while the nucleus of the atoms/molecules shift to the negative plate. This causes a difference in the locations of "center of charge" of electrons and molecules in the dielectric field. This difference create tiny displacement dipols (electric field vectors) inside the dielectric material. This field makes the free electrons in the positive plate go away, while it collects more free electrons to the negative plate. This is how charge is collected in the capacitor plates.

3.Am i correct in assuming that the resistor voltage drops because the capacitor's voltage is increasing? (kirchoff's law where volt rise = volt drop).

As the capacitor voltage increases, the voltage across the resistor will decrease accordingly because of the Kirchoff's Law, which I formulated above. So, yes, you were correct.

1.If the capacitor's voltage is dropping(due to it being discharged), shouldn't the resistor's voltage be increasing due to kirchoff's law? Also,this should therefore INCREASE the current instead of decreasing it, which would then cause the capacitor to discharge even faster?

You are missing the fact that, the source voltage is zero (i.e.; the voltage source is missing) in the discharge circuit. Substitude \$V_{supply}(t)=0\$ in the formula above. The capacitor voltage will be equal to the resistor voltage in reverse polarities during the discharge. Together, they will tend to zero.

I would not suggest deliberately directly shorting anything much bigger than 1000uF/16V ~= 0.1J (1/50 or so of your example), though I sometimes do it when I'm in a hurry and "pretty sure" 99.5% it's discharged through circuitry or whatever but would rather risk taking a nick from a screwdriver tip than my finger or meter. You could damage the shorting element or the capacitor doing that if it's fully charged. I have some fat 5W and 10W wirewound cement resistors around to do it properly, when necessary. If you misjudge the surge energy badly compared to what the resistor is capable of it's possible to have them explode or silently open up leaving the capacitor with a charge on it.

Very large capacitors can hold a lot of energy and demand serious precautions. Sometimes just the dielectric absorption can result in a big enough charge to give a significant jolt. That happens when you properly discharge a large capacitor and then remove the resistor- the capacitor appears to partially re-charge itself due to the way the dielectric behaves.