Resistance Being Proportional to Length And Its Relation to Magnitude of Current 
"Resistance of an electrical conductor is proportional to it length"

The intuitive explanation I found in many articles was that the greater the length of the conductor, such as a wire, the greater the number of collisions of the electrons with ions and therefore greater resistance. But how would this greater number of collision do anything to the drift velocity and therefore to the current $I$? Between any two collisions, the average speed of the electrons is the same and this affects the current? How would a greater length change this average speed between two collisions?
 A: Suppose we apply the same pd across a wire of twice the length. In that case the potential gradient halves, so the electric field strength halves. [This follows from $work=force \times distance$; the pd gives the work per unit charge; the field strength is the force per unit charge.]
If the electric field strength halves, the acceleration of the electrons between collisions halves, so the electrons' mean drift velocity halves. [The mean time between collisions isn't affected much, as the electrons' speed is almost entirely thermal, and far greater than the mean drift velocity due to the field. We make the crude assumption that on average each time it collides with the lattice, the electron loses all the velocity it has acquired from the field, and starts accelerating all over again.]
If the drift velocity halves, the current halves and the resistance doubles. 
A: The fact that resistance is proportional to length and inversely proportional to area can be intuitively shown with a bit of thought experiment.
Say you have a conductor of length $L$ and when you apply $5 V$, a current of $3 A$ flows through it. So the resistance is $\frac{5}{3} \Omega$. Now take another exactly similar conductor. Now if you join them end-to-end, you have a conductor of length $2 L$. Now if you have to pass $3 A$ of current through this new conductor, how much voltage do you need?
Well, since each of the conductors required $5 V$ to pass $3 A$ through them, it will take a total of $(5 + 5)V = 10 V$ across the combined conductor to pass the same amount of current. Hence the resistance becomes $\frac{10}{3} \Omega$, which is exactly $2$ times the original conductor. If you took $n$ conductors and joined them end-to-end, you would have a new resistance of $n$ times the original resistance i.e. resistance is proportional to it's length.
You could think of doubling (or $n$ times) the area in same way as before. Applying same $5 V$ across two conductors joined side-by-side will create $3 A$ current in each conductor, total of $6 A$ (or $3n\ A$)  through combined conductor. Making the new resistance being $\frac{5}{6} \Omega$, which is half (an n-th) of the original resistance. So resistance is inversely proportional to area.
Note: Now, you may say that this method just proves it for integer multiples of original length/area. To solve that, we could take help of differential lengths/areas. Let's consider the original conductor of length $L$ and area $A$. You could just say this conductor is actually stacked up vary tiny conductors of length $\frac{L}{N}$ and area $\frac{A}{M}$ where $M$ and $N$ are huge numbers. Rigorously speaking, we're talking about differential length and differential area. So now, you could just talk about any integer multiple of these differential conductors. That'd prove the proportionality for any length/area of the original conductor, not just the integer multiple ones.
N.B. What I did here is just simply thinking of increasing length as connecting more conductors in series increases equivalent resistance. And increasing area as connecting in parallel configuration which reduces equivalent resistance.
