I know this question has been left untouched for a while but I was thinking about this question myself and could not find any thorough examination of it, despite it apparently being as OP states quite a simple question. Nobody has yet addressed exactly why the waterfall analogy fails, only stating that it's not a 'good' analogy. On the contrary, I think it is a good analogy, after all, you would be hard pressed to explain the fundamental difference between the action of gravitational forces on water molecules in a waterfall and electric forces on charges in a circuit, without resorting to quantum mechanical arguments. So why does the analogy appear to fail?
Why does increasing the height of a waterfall not increase the (water) current, when increasing the potential difference in a circuit does increase the electric current?
I. Understanding the waterfall current.
First let us understand exactly what is causing the current in a waterfall. Water molecules passing over the top of the waterfall have zero downwards velocity (the velocity is tangential to begin with), but as they tip over the edge a force acts downwards (the gravitational force) which accelerates these molecules downwards therefore causing an increasing downwards current. However they don't accelerate indefinitely, because resistive forces acting (dependent upon velocity) mean that the molecules reach a terminal velocity (assuming the waterfall is tall enough for the molecules to reach this velocity). This terminal velocity $v$ is reached when the gravitational force equals the resistive forces. The 'current' of a waterfall is proportional to this velocity.
In paying attention to molecular causes of the current, we don't actually make any reference to $V$, the potential difference. This is because it is not the potential difference which causes the current, but the force $F$, which in our case is simply $mV/h$, where $h$ is the height of the waterfall and $m$ is the mass of a water molecule. This has to be the case - we know that <PERSON>'s Second Law tells us that forces cause accelerations (changes in $V$ over distances) rather than absolute changes in $V$. 780
The moral of the story is that we cannot only consider changes in $V$ when explaining the dynamics of a system, but must also consider the distance over which these changes occur, $h$.
Now say that we double the height of the waterfall. Then indeed the potential difference $V$ between the top and the bottom has doubled, but needless to say we have also doubled the height $h$! So in doubling the height of a waterfall, we have sent:
$$ V \mapsto 2V\ h \mapsto 2h. $$
Now we know the thing actually doing the work on the water particles is not the potential difference, but the force $F$, which in our case is:
$$F_{new}=m\frac{2V}{2h}=m\frac{V}{h} = F_{old}$$
Evidently we can see that doubling both $V$ and $h$ results in the same force, and therefore the same current. In other words, the thing determining the current of a waterfall (given it is tall enough for terminal velocity to be reached) is the gravitational field strength $g$, and changing the height of a waterfall does nothing to alter this.
II. The circuit current.
So much for the waterfall case, what about for circuits?
In making an analogy between the waterfall and the circuit, OP correctly points out that doubling the potential difference $V$ across a resistor of resistance $R$ doubles the current $I$, but fails to account for the doubling of the 'height' $h$. So suppose we double the potential difference across some resistor, but we also increase the 'height' of the resistor. What does doubling height mean in this context? In the case of the waterfall, $h$ is the total distance that the water molecules travel parallel to the field, and so the corresponding parameter in the circuit is the length $l$ of the resistor. Now doubling the length of a resistor in a circuit is the same as simply adding a second identical resistor in series. But by doing this we have doubled the resistance $R$! So we have sent:
$$ V \mapsto 2V\ R \mapsto 2R. $$
So evidently:
$$ I = V/R $$
stays the same! Paradox resolved.
III. Why the confusion?
I think that the core of the difficulty in understanding these analogies is that Ohm's Law appears to contradict <PERSON>'s Second Law - specifically it appears to not take any account of length scales! However of course it does take account of length scales - they are just hidden in the formula for the resistance $R$.
In both our situations, the mechanics of the individual particles is the same. 780