how not both points can be on the polygon implies that every point on the polygon is uniquely determined by its distance from given two points? I need help understanding the proof of lemma 2.1. in these notes here. The proof and lemma are the following:
Lemma 2.1. Every point on a regular polygon is determined, among all points on the
polygon, by its distances from two adjacent vertices of the polygon.
Proof. In the picture above, let the black dots be adjacent vertices of a regular polygon.
Since they are adjacent, the line segment connecting them is an edge of the polygon and
the polygon is entirely on one side of the line through the black dots. That shows the two blue dots can’t both be on the polygon, so a point on the polygon is distinguished from all
other points on the polygon (not from all other points in the plane!) by its distances from
the two adjacent vertices. 
I understand the proof up to "..., so a point on the polygon is distinguished from all
other points on the polygon..."

Please could someone help me understand how the fact that not both
  points can be on the polygon implies that every point on the polygon
  is uniquely determined by its distance from given two points?

 A: It means that, if one of the two points stays in the polygon the other stays not on the polygon, so, given the two radii $r_1$ and $r_2$ of the circles, there can be only one point in the polygon that has such distances from the two vertices, and this is true for all points in the polygon.
A: It's not hard to see that isometries preserve triangles so given the base of a triangle from $A$ to $B$, and a point $P$ that is not colinear with the segment defined by $A$ and $B$, call it $AB$ we can draw the segments $AP$ and $BP$ which will give us three lengths which completely define the triangle.
Now what happens if instead of considering $AB$ we consider some proper subsegment of it that does not contain $A$ or $B$ and call it $ab$. Then $aP$ and $bP$ define another triangle that must also be preserved, however if I slide $ab$ along the segment $AB$ towards one end or the other it will change the lengths of $aP$ and $bP$, so it will not be an isometry. That's why it's important to establish that $A$ and $B$ cannot both be on the polygon simultaneously without it being a full isometry and the extrema prove convenient for this purpose.
To see why consider the center point of the $n$-gon and the segments defining each side they form a triangle and since the $n$-gon is convex there is no overlapping triangles so we can partition the entire $n$-gon this way. Since the center of the $n$-gon is stabilized by the action and they are all the same length by regularity we only need one segment to characterize it.
Note too that we didn't actually need the regularity to get the partition the $n$-gon into triangles either, just the convexity. The same partition can be generalized when examining rigid motions on other convex shapes and in arbitrary dimensions.
