Why is the angle between the radial velocities at two instants the same as the angle between the tangential velocities at those same instants? 
While it is clear that the angle between $v_r$ and $v_r +  \Delta v_r$ is $\Delta \theta$, I cannot see a clear geometric reason as to why the angle between $v_t$ and $v_t +  \Delta v_t$ must also be $\Delta \theta$.
The book by Kleppner and Kolenkow uses this to argue that $\Delta v_t \approx v_r \Delta \theta$ and in the limit in which $\Delta t \rightarrow 0$ , $\frac{d v_t}{dt} = v_t \dot{\theta} = r\dot{\theta}^2 $.
 A: If I'm correct, this is on p37, also this book is a classic, so many people have it, referencing it properly would speed things up.
First, I'm assuming that Vt is the component of velocity tangent to the f(x) that is your objects position, and Vr, well is just radial velocity.
Before diving in, I want you to show how your understanding is wrong of the problem on hand. Vr is not always perpendicular to tangent of a curve, rather the normal to that curve is that, radial is meant as anything to do with respect to the center of one's coordinate system, be it distance, velocity or others. Now you might say that the tangent is to the curve, rather than the conventional meaning (theta direction in the polar coordinates you mentioned).
Let's assume that there is a strait line, say with equation x = const., so if Vt is tangent to the line of motion, it's always the same.
So let's assume that the tangential component, is in fact the component perpendicular to the radial velocity, so the overall speed is root square of the sum of 2 velocities, written as:
$$V(total)^2 = Vr^2  +  Vt^2$$
Now, if you think of your tangent and radial velocity like the figure below, everything will fall into place:

A: The statement is obviously true for motion in a circle.  For any short arc (swept out in a short time) a center of curvature can be found so that the arc can be considered part of a circle.
