>If we think of T as a curve of its own, then N=T' / ||T'|| should literally calculate the tangent vector to T, no?

Yes, precisely.

>And even if we think of T as being the tangent vector to some curve r, this would mean that N is comprised of the second derivatives of r, which still doesn't make intuitive sense because in 2D we don't get perpendicular lines by differentiating twice.

Yes, the normal vector is related to the second derivative of the original curve.

You can't quite say that N=r''. You're right that r'' could point in any direction, but you can always decompose it into a tangential component plus another component orthogonal to that. The tangential component of the acceleration tells you how much you are speeding up or slowing down: it is a force acting along the direction of movement, so it is adding or removing energy in the usual work=force\*distance way.

The other component is the normal component of the acceleration, which causes your direction to change but does zero work. The unit vector in this direction is the standard normal vector N.

For example, the normal vector to a circle points toward the center of a circle. If you think of that circle as an orbit, it should make sense: gravity (a force, which causes acceleration, ie the second derivative) is pulling you towards the thing you are orbiting, which is at the center of the circle.

Also, to revisit this line again:

>in 2D we don't get perpendicular lines by differentiating twice.

You're right: you get perpendicular lines by differentiating *once*! I mean, not always of course, but certainly for something like T that satisfies |T|=1. Think about why this is. Hint: if T' was *not* perpendicular to T, then T' would have a nonzero tangential component...