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're right. I'll work my way up to an answer. First, for current to flow, you need a force to push charges around the loop. The force that does this is the electric field. There's an electric field within the wire and it follows the shape of the wire around the circuit. This electric field is responsible for giving the charges of the wire a net drift direction ("Friction" is what keeps the pushing force from accelerating the charges to infinity - so we obtain a nice average drift speed of charges).
$E$ Field in Conductors
There's nothing wrong with an electric field in a conductor. We usually think that electric fields have to be 0 within a conductor, but this is only in the static case. Imagine applying an electric field to a conductor. You probably already know this. But the electric field is definitely not zero in the conductor. It's only zero after everything has settled down and we are in the regime of electrostatics. However, before we got to electrostatics, there was definitely an $E$ field in the conductor. Likewise for circuits, the wire is a conductor but it is never able to reach electrostatics (the details of which are a little nuanced - the battery essentially prevents the wire from reaching a static situation). Therefore having an $E$ field within the wire is completely fine. Where does this $E$ field come from? This is getting a little bit off topic but the battery has an electric field. It's this electric field plus the electric field of induced/piled up charges along the surface boundary of the conducting wire that shapes the field within the wire. Anyways, the point is that an $E$ field exists and does the pushing. There's one other thing you have to know: for many substances $J = \sigma E$ where $J$ is current density and $\sigma$ is conductivity. This is Ohm's law from which $V = IR$ can be derived. For copper, $\sigma$ has an order of magnitude of $10^7$. The point is that $E$ is very small in a conductor.
Potential $V$
Wires do have a potential drop $V = - \int \vec{E} \cdot d\vec{l} < 0$ as $\vec{E}$ isn't zero and points in the same direction as $d\vec{l}$ (assuming we are integrating in that direction). So start at terminal $a$ of the battery and move to the start of the resistor, point $b$. The voltage drop is $V(b) - V(a) = - \int_a^{b} \vec{E} \cdot d\vec{l}$, which is miniscule because $E$ is so small. In the resistor, the electric field $E$ becomes really large (low conductance $\sigma$). Therefore the voltage drop $V(c) - V(b) = - \int_b^{c} \vec{E} \cdot d\vec{l}$ is large because $E$ is large. Then on the other side of the wire, because you still have your $E$ field pushing, you'll have a tiny voltage drop again. These voltage drops in the wire can be neglected.
Best Answer
It is a line for most intents and purposes.
If you dig much deeper there are countless effects all of which make it slightly nonlinear. They could be getting at all sorts of effects.