How do voltage and voltage drops over a circuit relate to work done?
The Volt unit is energy normalized to unit charge; Joule per Coulomb.
Since the Amp unit is Coulomb per second, the product of the voltage across and current through a circuit element is the power associated with the circuit element.
For a DC circuit, voltage and current are constant thus the energy delivered or supplied by a circuit element over some period of time is the product of the voltage across, the current through, and the elapsed time.
Now, the rest of your question contains misconceptions that are too numerous to untangle here. I recommend that you do some more reading and careful thinking about, in particular, basic circuit laws. For example, Ohm's Law immediately gives you the answer to your question "Does voltage drop over a resistor decrease as current decreases?"
Ohm's law states that the current through a conductor between two
points is directly proportional to the potential difference across the
two points.
Also, I recommend perusing William J Beaty's "Electricity Misconceptions Spread by Textbooks" site.
For a start, take a look at Electric current is a flow of energy? Wrong.
In general it seems that if one measures the voltage output before
some component the measurement will be equal to the drop of voltage
accroos the said component.
In this case, the voltage $V_2$ is a node voltage which means that is the voltage between the output node and the ground node. By definition, the ground node voltage is zero
$$V_0 = 0 \mathrm V $$
But, by inspection, the voltage across the resistor $R$ in the left-most circuit is simply
$$V_R = V_2 - V_0 = V_2 - 0 = V_2$$
Similarly, the voltage across the capacitor in the right-most circuit is simply
$$V_C = V_2 - V_0 = V_2 - 0 = V_2$$
Keep in mind that the output voltage must be taken across two nodes of a circuit. If there is just one circuit element connected between the two nodes, the output voltage is simply the voltage across that circuit element.
However, there may be a complex network connected between the output nodes so it isn't generally true that the output voltage is just the voltage across a single circuit element.
Best Answer
When two resistors $R_1$ and $R_2$ are in series, the total resistance is:
$$R=R_1+R_2$$
If we apply a voltage $V$ across both resistors, the current $I$ is given by:
$$I=\frac{V}{R}=\frac{V}{R_1+R_2}$$
The voltage drop $V_i$ (for $i=1$ or $i=2$) across each resistor $R_i$ is:
$$V_i=R_iI=R_i\frac{V}{R_1+R_2}$$
So the voltage drops are not the same if $R_1\neq R_2$.
What determines the brightness of a bulb $i$ is its power output, $P_i$: $$P_i=V_iI=R_i\frac{V^2}{(R_1+R_2)^2}$$
So in series, the power of each bulb is proportional to each resistor $R_i$.