Why $V\propto R$ and $V\propto I$? Ohm's law's gives us two relations: $V\propto R$ and $V\propto I$. But why is that so?
Potential difference($V$) between two points depends on the concentration of charges between them. Current flows from a region of high concentration of electrons to low concentration of electrons(low to high potential). But I can't seem to figure out how $V$ is dependent on:

*

*Current($I$): According to my book-


Potential difference between two points is equal to the work done per unit charge in moving a positive test charge from one point to another.

So why does current, which is number of charges flowing per unit time a factor in the work done on unit charge? If the work done on a unit charge to bring it from one point to another is $x$ then the work done to move $5$ charges between the same points is $5x$, but the potential between the two points, as per the definition, remains $x$.

*

*Resistance($R$)-
Suppose there is unit positive charge Q which is being taken to a greater positively charged body.

Suppose external work done is $x$ and then some resistance is provided in the path of the charge.
We know that $$\Delta W_{external}=\Delta PE + \Delta KE + W_{other}$$ where $W_{other}$ comprises other forms in which energy is being lost, e.g., heat, against friction, etc. So isn't work done against resistance coming under $W_{other}$? Just like in gravitational potential energy, where if we lift a body up a certain height through air, then stop the ball and keept it at rest at that certain height, then $W_{external}=\Delta PE+ \Delta KE + W_{againstresistance}$. $\Delta KE$ is $0$, and the total external work done is $\Delta PE + W_{againstresistance}$, but the change in potential always remains $mg \Delta y$!
Then why is the potential difference not only dependent on the shortest distance between and the charge concentration of the two points, but instead on $I$ and $R$?
 A: For many physics problems the way to solve them is to get the right level of abstraction. You have drifted into the wrong level.
You need to be careful about comparing currents in wires to motion of particles affected by potential differences on static charges. Your diagram shows a test charge being moved from infinity towards a fixed charge. Electric circuits are not like that. Ohms law is about circuits.
As a snap-you-out-of-your-rut consideration, consider an electron going around a circuit several times. It moves "up" the resistance each time. Does it therefore wind up at a higher potential each time? Or does something happen so that it gets back down the hill each time?
The potential difference between the leads of a resistor determine the energy a single electron will dissipate in the resistor as it goes through. It does not gain that much energy. On the other side of the resistor it will have pretty much the same energy it had in the wire before the resistor. It will be stumping along at the same speed it had before.
The electron is getting pushed through the resistor. The restance is using up the energy of that push, mostly as heat. Tiny bit of electromagetic radiation.
When you increase the potential, increasing V, more current flows. And each electron loses more energy passing through the resistor. Double V you also double I, and you multiply the power into the resistor by 4. They do go faster, but the kinetic energy in an electron is quite small compared to the energy the current is carrying.
So your diagram of bringing a test charge from infinity is misplaced. You want to think of the electrons on a string getting dragged across rough surfaces. They don't wind up with more energy even if you pull harder. You just get more per second going by any given spot. And by pulling harder each one gets more violently smacked around by the rough surface.
A: If you are looking for an more intuitive understanding then maybe look at 'cause and effect'
A circuit with 0 voltage at any point evidently has no current flowing.
A potential difference provides a 'push' (the water pressure analogy DOES work here) to make the current flow.
It seems intuitive that the harder you push the more will flow. Ohms law simply formulates that the relationship is linear.
So rather than think of  V∝I it is more the case that  I∝V.
Whilst they are mathematically equivalent the second suggests more clearly that current is caused by the PD.
A: Voltage - energy required to bring a charged particle from infinity to a specific field/region.
Mathematical formulation : $$V=-\int_\infty^a \vec E\cdot \mathrm d\vec r$$ In circuit- we take two points between a resistance(considering the circuit is series circuit). Cause there's some energy lost for resistance. Resistance converts some sort of energy to heat. $$P=I^2R$$ to derive the equation you can apply two equations $$P=\frac{W}{t}\tag{1}$$ $$W=VQ$$
Current- amount of charge flow in a specific region per sec(the definition in your book is also correct)($I=V/R$). Math :$$I=\frac{q}{t}$$
In 18th, 19th or 17th centuries an experimentalist had done a work with flow of current. He observed as, voltage increases the rate of charge flows also increases. For different materials the value was different, he had drawn a graph potential vs current. he found straight line for all materials. That's why he thought $$I\propto V$$. Further he thought something is stopping flow of charges which was called resistance and since it is decreasing flow of charges that's why resistance is inversely proportional to current $$I=V/R$$
