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.