But Where Does it Start?

Relevant wiki: Regular Polygons - Properties of Regular Polygons

A segment is formed by joining any 2 vertices.
Thus total segments $\dbinom{8}{2} = 28$
Out of these 8 segments are the side and remaining are diagonals.
Thus total diagonals = 20.
For an n-gon the total number of diagonals = $\dfrac{n(n-3)}{2}$

We have to choose two points from 8 points and connect them together which can be obtained by ${\phantom{0}}^8C_2 = \dfrac{8!}{2!}{6!} = 28$ . Now, 8 of these lines are the sides of the octagon. So, the answer is $28 - 8 = \color{#69047E}{\boxed{20}}$ .

Generalizing:-
The number of diagnols in a n-sided polygon $= {\phantom{0}}^nC_2 - n\\ = \dfrac{n!}{(n-2)! × 2!} - n\\ = \dfrac{n(n - 1)}{2} - n\\ = \dfrac{n^2 -n - 2n}{2}\\ = \dfrac{n^2 - 3n}{2}\\ = \dfrac{n(n - 3)}{2}$

Let us name the vertices A,B,C,D,E,F,G,H. There are 5 diagonals from each vertex(because itself and the 2 adjacent vertices are not counted). 8 vertices * 5 = 40 diagonals. But, you counted each diagonal twice, thinking diagonal AE is different than diagonal EA! So,40/2=20 diagonals.

Number of diagonals of polygon of n sides = n(n-3)/2.