Is there a reason why curvature is defined as the change in $\mathbf{T}$ with respect to arc length $s$ And not with respect to time $t$? (or whatever parameter one is using)
$\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{dt}}|$ seems more intuitive to me. 
I can also see that $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{ds}}| = |\frac{d\mathbf{r}'(t)}{dt}|$ (because $\displaystyle |\mathbf{r}'(t)| = \frac{ds}{dt}$, which does make sense, but I don't quite understand the implications of $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{dt}}|$ vs. $\displaystyle |\frac{d\mathbf{T}(t)}{\mathit{ds}}|$ and why the one was chosen over the other. 

The problem is that if you define it in the "more intuitive way" the curvature depends on the parametrisation. We calculate the curvature of a curve, not of a parametrisation.
Simple question: how do you calculate the curvature of the hyperbola $y^2-x^2=1$ at let's say $(0,1)$?
You could 
a) solve for $y$ and use $x=t, y= \sqrt{t^2+1}$.
b) use $x= \tan(t), y= \sec(t)$,
c) use $y= \cosh(t) , y= \sinh(t)$.
Each of these lead to a different $\displaystyle \left\vert\frac{d\mathbf{T}(t)}{\mathit{dt}}\right\vert$. So which one would you pick as the curvature of $y^2-x^2=1$?
Intuitively, when we parametrise a curve, we are basically describing the curve as being the trajectory of a particle, by describing its coordinates at time $t$. I always think about different parametrisations as being different particles moving on the same curve/trajectory.
If a particle moves faster, the calculation of $\displaystyle \left\vert\frac{d\mathbf{T}(t)}{\mathit{dt}}\right\vert$ also takes into acount its velocity. If I am not mistaken, in the arc lenght parametrisation, we simply pick the particle which covers 1 unit of the curve per unit of time. Basically we decide that the speed is constantly 1. This is the most "natural" choice you can make.
