According to Ohm's law, the voltage across and current through a resistor are proportional.
If one places a variable resistance (rheostat) across a voltage source, the voltage across the rheostat is fixed.
By fixing the voltage across the rheostat, the current through becomes inversely proportional to the resistance:
$$i(R) = \frac{V}{R}$$
A circuit with just a voltage source and rheostat is uninteresting. Consider adding a light bulb in series.
With the rheostat set to $0 \Omega$, the light bulb produces full brightness.
The resistance of the bulb and rheostat are added in series. And, as before, the voltage across the series combination of rheostat and bulb is fixed.
Thus, as the resistance of the rheostat increases, the current through the bulb decreases.
However, we can also say, with equal validity, that the rheostat controls the voltage across the bulb.
If $R_P$ is the resistance of the rheostat and $R_B$ is the resistance of the bulb, the voltage across the bulb is given by voltage division:
$$V_B = V \frac{R_B}{R_P + R_B}$$
If this were the electrical engineering stack exchange site, I could add easily add a schematic of the circuit for clarity. Perhaps this question should be migrated.
You said in your question that you knew that the potential energy per charge was the volatage. One way of writing this is that the potential energy lost per unit time (power dissipated, $P$) is the charge flowed per unit time (current, $I$) times the potential $V$. Therefore we get $P=IV$ or $V=P/I$. From this equation we see that the potential is indeed inversely proportional to the current assuming constant power.
Now in "real life" we usually don't have constant power. For example, let's assume I have light bulbs of two different resistances $R_1$ and $R_2$ (let's say with $R_1 < R_2$), and I plug them both into the wall. The way electricity from the wall works is that it supplies a constant voltage $V$. Now we expect that with constant $V$, the light bulb with a lower resistance will allow more current to flow. In fact, the current $I$ resulting from a voltage across a resistance is given by $I=V/R$. Therefore we will get $I_1=V/R_1$, and $I_2=V/R_2$. Since $R_1$ is smaller, its current $I_1$ will be bigger and so the power $P_1 = I_1 V$ will also be bigger. Since the power is not the same for the two light bulbs, we do not get a contradiction with your equation from the first paragraph.
Best Answer
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.