Choose one to be held constant.
Suppose we supply the circuit with a constant power source, for example a $9$ $V$ battery..
If you double the resistance on the circuit, the current drawn from the power source will halved.
Suppose we changed out the $9$ $V$ battery with a $18$ $V$ power source while holding the resistance constant, the current will double.
There isn't a chain reaction like you think would happen. Everything happens in "one-step". Voltage will always reflect the current and resistance at that precise moment.
Just because I add more resistance, the loss in current isn't going to change my $9$ $V$ battery to a $4.5$ $V$ Battery.
The current is halved BECAUSE it's still a $9$ $V$ battery with doubled resistance.
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
If you keep the resistance constant, then $V=IR$ means that voltage is directly proportional to current.
If you keep the power constant, then $V=\frac{P}{I}$ means that voltage is inversely proportional to current.
However, because $V=IR$, we can write that $P=I^2R$. Therefore, if we say resistance is constant, then power must change with current, which means that voltage is no longer inversely proportional to current.
There is no contradiction here, you simply need to be mindful of what you are holding constant and ask yourself if you are being consistent