You consider the example f'(x)=x^2, but notice how this function does not satisfy |f'(x)|=1.

Only when |T(t)| is constant can you be sure that T(t) and T'(t) are perpendicular (in particular, this is true for |T(t)|=1 for all 1).

For an easy example, take the circle parametrized by r(t) = (cos(t),sin(t),0) (or you could leave out the last coordinate, if you are comfortable with plane curves).

Certainly |T(t)|=1. Also, T(t) = (-sin(t),cos(t),0), while T'(t) = (-cos(t),-sin(t),0). Notice how T(t) and T'(t) are perpendicular.