Please explain how Richardson Extrapolation is used in this example I am having a hard time understanding what on earth Richardson Extrapolation is trying to do.

Consider the example of approximating $\pi$ by inscribing regular polygons in a unit circle.
The perimeter of a regular $n$-gon is $2n\sin \bigl(\dfrac{\pi}{n} \bigl)$, whereas the circumference of a unit circle is $2\pi$, so that $2n\sin \bigl(\dfrac{\pi}{n} \bigl) \approx 2\pi \implies n\sin \bigl(\dfrac{\pi}{n} \bigl) \approx \pi$. So let $c_n = n\sin \bigl(\dfrac{\pi}{n} \bigl)$.
If we put $n=\dfrac 1h$, then
$$c_n=\frac 1h \sin (\pi h)=\pi - \frac{\pi^3h^2}{6}+\frac{\pi^5h^4}{120}+\cdots$$
With a bit of trigonometric manipulations, we get the recurrence relation
$$c_{2n} = 2n\sin \biggl (\frac{\pi}{2n} \biggl)=2n\sqrt{\frac 12 \biggl(1-\sqrt{1-\Bigl(\frac {c_n}{n}\Bigl)^2}\biggl)}$$
so $c_2=2$, $c_4=2.8284$, $c_8 = 3.0615$, $c_{16} = 3.1214$ and so on.
I still understand what is happening up to here, but then it suddenly says, without any explanation,

Extrapolating between $c_4$ and $c_8$, we get $c_4^{(2)}=3.1391$; and between $c_8$ and $c_{16}$, we get $c_8^{(2)}=3.1214$.

Can anyone explain to me why we get these numbers?

Apparently, there is the formula
$$T_h^ {(2)} = \frac{4T_{h/2}-T_h}{3}$$
where $T_h$ is an approximation to $T$, and $T_h^{(2)}$ is a "more accurate formula".
But plugging in the numbers doesn't seem to work:
$$c_4^{(2)}=\frac{4c_2-c_4}{3}=\frac{4(2)-(2.8284)}{3}=1.7239 \neq 3.1391$$
 A: Richardson extrapolation is often phrased in terms of a stepsize $h$ which is small. Here I will phrase part of the theory in terms of an integer $n$ which is large.
In your case of the cirle, you have
$$ c_n = n \sin\left(\frac{\pi}{n}\right) = \pi - \frac{1}{3!} \frac{\pi^3}{n^2} + \frac{1}{5!}\frac{\pi^5}{n^4} + O(n^{-6}).$$
It follows that
$$ \pi - c_n =  \frac{1}{3!} \cdot \frac{\pi^3}{n^2} - \frac{1}{5!} \cdot \frac{\pi^5}{n^4} + O(n^{-6}).$$
This is an example of an asymptotic error expansion. By replacing $n$ by $2n$ we discover that
$$ \pi - c_{2n} = \frac{1}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2} - \frac{1}{16} \cdot \frac{1}{5!} \cdot \frac{\pi^5}{n^4} + O(n^{-6}).$$
We would like the ability to estimate the dominant error term, i.e., the term
$$ \frac{1}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2}.$$
By subtracting $\pi-c_{2n}$ from $\pi-c_n$ we find
$$ c_{2n} - c_{n} = \frac{3}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2} - \frac{15}{16} \cdot \frac{1}{5!} \cdot  \frac{\pi^5}{n^4} + O(n^{-6}).$$
It follows that
$$ \frac{c_{2n} - c_n}{3} = \frac{1}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2} - \frac{5}{16} \cdot \frac{1}{5!} \cdot \frac{\pi^5}{n^4} + O(n^{-6})$$
We can suppress some information and write this as 
$$ \frac{c_{2n} - c_n}{3} = \frac{1}{4} \cdot \frac{1}{3!} \cdot \frac{\pi^3}{n^2} + O(n^{-4}).$$
Observe, that the number on the left hand side is an approximation of the dominant error term. Moreover, it can be explicitly computed.
By combining past expressions we have
$$ \pi - \left( c_{2n} + \frac{c_{2n} - c_n}{3} \right) = O(n^{-4}).$$
In short, if $c_{2n}$ is a good approximation, then (presumably) we can construct a better approximation using the formula
$$ c_{2n} + \frac{c_{2n} - c_n}{3}.$$

Do not subcumb to the temptation to write or compute this as
$$ \frac{4 c_{2n} - c_n}{3}. $$
The two formulae are equivalent in exact arithmetic, but the first formula is superior when using finite precision arithmetic.
A: You have your formula backwards.  Remember,  the c_n with the larger n multiplies by the higher power of 4.
You need to change $$ c_4^{(2)}=\frac{4c_2-c_4}{3}=\frac{4(2)-(2.8284)}{3}=1.7239 \neq 3.1391$$
to $$  c_4^{(2)}=\frac{4c_8-c_4}{3}=\frac{4(3.0615)-(2.8284)}{3}=3.1392$$
