I think you've answered that yourself. If you are putting more work into moving unit of charge, then that unit of charge is going to move faster (all else being constant). Current is the flow electric charge across a surface at specific rate (1 ampere = 1 coulomb per second) and hence - more voltage, more work, faster flow (rate), higher current.
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.
Best Answer
As usefully pointed out in the comments, Ohm's law just states that voltage $V$ and current $I$ have a relation. Therefore, Ohm's law per se means equivalence and not causation. This could be not immediate to deduce, there is plenty of Physics laws which imply both a relation and a causation, but this is not the case for neither Ohm's law, nor Newton's second law, as you correctly mentioned.
The causation is established according to the specific circuit and its features.
For example, a battery is a device which tries to always impose a fixed voltage $V_b$ between its terminals $V_+$ and $V_-$. In this example, $V_b$ is the cause and the independent quantity. If you connect a lamp between the battery terminals, it can be represented as a resistance $R_{\mathrm{lamp}}$. The voltage across $R_{\mathrm{lamp}}$ is $V_b$. The current on $R_{\mathrm{lamp}}$ is then the effect of the presence of the battery, and the dependent quantity. This is a little circuit which consists in a voltage generator (the battery), a resistor (the lamp) and a single loop. As an effect of the imposed voltage, a current $I$ will flow through the resistor. $I$ is the dependent quantity: to determine its value, you must know both $V_b$ and $R_{\mathrm{lamp}}$; Ohm's law can be written as
$$I = \frac{V_b}{R_{\mathrm{lamp}}}$$
Of course, the battery may not be able to erogate enough current to create or maintain a $V_b$ across $R_{\mathrm{lamp}}$, or the current may be so high that $R_{\mathrm{lamp}}$ is damaged, and so on, but these real cases are not relevant here.
On the other hand, transistors can be used as current generators in several configurations. Current generators try to always impose the value $I_b$ of current between their terminals $I_+$ and $I_-$. If you connect a resistor $R_{\mathrm{lamp}}$ between these two terminals, you already know the current $I_b$ flowing into the resistor: $I_b$ is the independent quantity and the cause. This is again a simple circuit consisting in a current generator, a resistor and a single loop. As an effect of the current $I_b$, a voltage $V$ will appear across the resistor: in this case, $V$ is the dependent quantity. To determine its value, write Ohm's law as:
$$V = R_{\mathrm{lamp}} I_b$$
There are again some real scenarios when $I_b$ can not be imposed by the current generators, but they can be discarded here.
The same Ohm's law is used in both the examples, but with exchanged roles between $V$ and $I$, according to the circuit.
In any circuit, regardless of its structure, if you know the value of a resistor $R$ and the current $I$ flowing through it, you can determine the voltage between the resistor terminals as $V = RI$. If instead you know the voltage existing between the resistor terminals, you can determine the current as $I = V / R$. In the previous cases, the cause was always the independent quantity and the effect was the dependent quantity. In this case instead, you don't know what is the cause, but you can still use Ohm's law: the dependent quantity is now, for you, the one you still do not know.
As correctly pointed out by Whit3rd's answer, Ohm's law mainly refers to a resistor in a circuit and it involves three quantities: the voltage across it, the current through it and the resistance value. If you know two of them, regardless of what is the cause and what the effect, you can determine the third one.