Time derivative of the area enclosed by a loop. I'm looking for some clarification on what the author means. 
One really has to take a look at the context (the image I posted is from page 23, so it's not much work) before answering my questions, but here they are: What does it mean for a double point to be transversal (and how does the first claim follow?)? How does the author compute the time derivative at time 0 of the area enclosed by the loop and gets to that integral (and evaluates it)? 
 A: A double point being tranversal means that the curve isn't tangent to itself. An example of a non-transversal double point would be here: Figure 1: the curve intersects itself in the middle, and is tangent to itself there as well.
2.13 is a set of first-order equations. Therefore, the initial conditions consist of the values and first derivatives. Uniqueness of solutions means that given a particular set of initial conditions, there is only one solution. 
If a curve crosses itself, then there are two different solutions at that point (if you start at that point, there are two different portions of the curve, corresponding to two different solutions.)  The values are the same at that point, so the derivatives have to be different. Imagine a car driving along the surface, and each point on the surface is affecting curve, making it turn. Suppose the resulting curve looks like this: Figure 2. If the car comes in from the bottom left, it will continue on to the upper right. It won't go to the bottom right or top left, because those would require it to have a different velocity. But in Figure 1, a car approaching the middle from the bottom left would have the right velocity to continue either to the top  right or bottom left; there isn't enough information to distinguish between the two.
I'm not sure where the integral comes from, but evaluating it as $-(\pi +\alpha)$ comes from noting that it is the turning number of the loop, excluding the point of intersection. 
