Arc Length parametrization significance For a curve $r=r(t)$, its arc length can be written as $ds = v(t) dt$ where $v(s) = |r'(t)|$. If $t=s$ then $ ds = v(s)ds$ giving $v(s)=1$. What's represented by the quantity $v(s)$? What's the siginicance of $v(s)$ being equal to $1$ and what's the intuitive explanation behind its constant value of 1?
Secondly, why do we do the arc length parametrization? Is there a reason that's unique to it?
 A: The definition of curve on $\mathbb{R}^2$ generally is achieved by  a suitable map $r: [a,b] \to \mathbb{R}^2$, that is, for each value of the parameter we identify a point of $\mathbb{R}^2$.
A curve is then essentially a set of points, to start with.
It is suggestive to think of the parameter as physical time: then the map above tells you where to be at a given moment of time.
It is then intuitively clear that many parametrization could describe the same curve: in a physical interpretation, they correspond to walking along the curve at different speeds.
Given a curve, I could walk along it at uniform speed, or accelerate and decelerate, and so on.
The key point is the length of the curve must not depend on such detail, in exactly the same fashion we define the length of a road without considering the speed at which people travel along it.
On the same physical grounds, we expect to be able to find the length somehow as $$ \int_a^b v(t) \mathrm{d}t$$ as the infinitesimal length travelled $\mathrm{d}s = v \mathrm{d}t$, where $v$ is the instantaneous velocity.
The velocity turns out to equal $\vert r(t)’ \vert$ as you say.
In a case such as
$$
r(t)  : [0;2\pi]\to
\left\{
 \begin{array}{ll}
  x(t)   = \cos(t) \\
  y(t) = \sin(t) 
 \end{array}
\right.
$$
the velocity equals $v = \cos^2 t + \sin^2 t = 1$, we are describing a point going around a circumference at unitary speed.
We could also have used
$$
r(t)  : [0;\pi]\to
\left\{
 \begin{array}{ll}
  x(t)  = \cos(2t) \\
  y(t) = \sin(2t) 
 \end{array}
\right.
$$
And this would represent a point rotating at double the speed, as an easy check confirms.
Both would give the same length when integrated over the proper interval of course.
Yet, if you are fortunate or cunning enough to work with a parametrization for which the velocity $v = 1$, your length integral is much simpler, and it simplifies immediately to $b-a$ (clearly the length equals time $\times$ speed, at the latter is one). 
This is the case you mention: if time equals length (your $t = s$, the speed must be equal to 1!).
If the speed is not constant for a different parametrisation, the length will not change, but you might have to solve a more difficult integral. 
You could check by yourself by substituing a function in the arguments of the $\sin$ and $\cos$ functions above: try to find a function $f$ such that $f(0) = 0$ and $f(2\pi) = 1$, so that the starting and ending points of the curve coincide.
EDIT following the OP's request in the comments
As I tried to hint at $v(s) =1$ is not a general requirement: it is something nice to have, something that can be arranged, but it is not a general property of all parametrisation.  I will try to clarify with a simple  example. 
Consider the segment $[0,1]$,  of length $1$. It can be described by, among others, any parametrisation of the type
$$ x_k(t) = t^k$$ for $t \in [0;1]$. For any of these, $x_k(1) = 1$ so they will end at the same point, which means, they will have travelled the same distance.
Yet there is one, for the case $k=1$, which describes a point moving at unitary speed, $v=1$, as its derivative is constant.
$v=1$ occurs only when we choose a very convenient parametrisation, arc-length, corresponding to a point moving along the curve at constant speed and covering over any interval a distance equal to the employed time. 
If the travelled  adistance equal the time employed for the trip, whatever time you look at, the speed will trivially be equal to 1.
$s = t$ imposes the condition $v=1$, because $v = \frac{\mathrm{d}s}{\mathrm{d}t}$.
If I am missing something, as I fear, do not hesitate to underline it and I will clarify more if I can.
