What you could say is "if energy is lost in a resistor, then why doesn't the velocity of the charged particles increase, as per Work-Energy theorem?". So your question should really go something like "Why doesnt current which is $Q/t$ increase if the velocity will increase after voltage drops"?
The answer to this is that we assume all potential energy lost, is again lost due to inner collisions with other atoms, and that's why materials heat up! Also, this implies a steady current because the drift velocity will not be changing.
Edit:
It can be shown that $F\Delta x= \Delta KE$ this means that a force acting on some distance will produce a change in kinetic energy. You can imagine an electric field where the work done by it is simply $W= qE\Delta x$. By the way notice that if I were to divide that expression by the charge $q$ I would obtain the voltage across the thing i'm concerned with. Namely $V=E\Delta x$.
So in summary, if I do some work, then I have some change in kinetic energy. If there is a voltage drop, it follows that positive work was done and kinetic energy increased, which means a velocity increment.
Across a resistor there is a voltage drop. So why isn't it that charged particles are going faster and then I can measure a current increase? Well that's due to the explanation I gave above my edit portion. Namely, that all that increase in kinetic energy is absorbed due to collisions with the neighboring atoms.
By the way, if you're curious enough to visit this website, I suggest you look up a video on Work-Energy theorem on youtube. This concept is pretty straightforward and I'm sure you can understand it.
You may have found a small glitch in that water fall analogy. An analogy I like much better is to think of water through pipes.
The voltage (potential difference) corresponds to the pressure difference between two points. A higher pressure in one spot means a larger "push" on the water. For charges in a circuit, the voltage is the "push" that squeezed them forward through the obstacles in the form of resistors and other circuit components.
Such a pressure difference is directly corresponding to a larger potential energy difference. This is why the water fall analogy is often used, because it is a more intuitive way to think of potential energy. But when you are increasing the voltage across two points in a circuit, then this corresponds to not a higher "pressure" difference from the top to the bottom of the water fall, but rather to a larger potential difference. And such a higher potential difference means a higher water fall, because the potential energy we are comparing with here is gravitational.
So, the increase in height of the water fall is analogous to an increase in charge accumulation in an electric circuit. The distance is changing in that analogy so the speeds are not really comparable. But they would be in the pipe-analogy.
Best Answer
A voltage source is assumed to deliver energy with a specific terminal voltage which does not depend upon the current from the source. A current source on the other hand is assumed to deliver energy with a specified current through the terminals.
Both current and voltage sources are ideal. In practice, we represent a real voltage source as an ideal voltage source in series with a resistance and a real current source as an ideal current source in parallel with a resistance.
All the best.