Understanding the unit tangent vector The vector $\dot x (s) $ is called the unit tangent vector to the oriented curve $x=x(s)$.
I am told that $x=x(s)$ is a natural representation of a regular curve C. 

What does natural representation mean?

The derivative $\dot x(s)=\frac{dx}{ds}$ is defined as the tangent direction to  C at the point $x(s)$. 

If $x=x(s)$ is a representation of a curve, how can it be a point?
Why does the length of $\dot x(s)=1$ ?
What does it mean to say the quantity $\dot x(s)$ is an oriented
  quantity?


$s$ is the arclength. Namely, (and we just work in $\mathbb R^2$ here):
if $\gamma:[0,1]\to \mathbb R^2$, is a paramterization of your curve $C$, then 
$\gamma (t)=(x(t),y(t))\in C$ and $\gamma(0)=(x_0,y_0); \ \gamma(1)=(x_1,y_1)$
then the length of $C$ from $(x_0,y_0)$ to $(x_1,y_1)$ is given by 
$l=\int_{0}^{1}\sqrt{\dot x^2+\dot y^{2}}dw$ and this is just $ \int_{0}^{1}\left \| \frac{d\gamma}{dw} \right \|dw$.
This formula is dervied easily from the Riemann Sums that arise from considering polygonal approximations to $C$.
Generalizing this, i.e. we have the function
$$\tag1l(t)=\int_{0}^{t}\left \| \frac{d\gamma}{dw} \right \|dw,\ $$ which measures the length of $\gamma $ from $\gamma (0)=(x_0,y_0)$ to $\gamma (t)= (x,y)$.
Now if we set $s=l(t)$ then $t=l^{-1}(s)$ which means that we can define a new parameterization $\overline \gamma (s)=\gamma (l^{-1}(s))$. That is $$\tag2\overline \gamma =\gamma \circ l^{-1}$$
$\overline \gamma $ is what is meant by the natural representation of a regular curve $C$, and here is why:
if we use $s$ in (1) with $\overline \gamma $ we obtain $l(s)=\int_{0}^{s}\left \| \frac{d\overline \gamma}{dw} \right \|dw$. But $l(s)=s$ because a change in parameterization does not affect the length of the curve (why?), and now using FTC we have $1=\frac{ds}{ds}=\left \| \frac{d\overline \gamma }{ds} \right \|$ and so we see that $\overline \gamma $ is the parameterization that makes the rate at which the curve is traversed $constant$ and equal to $1$. You can show that it is unique with this property. 
Another way to see this is the following:
If we differentiate $\overline \gamma $ wrt to $s$, we get: $\overline \gamma'(s)=\gamma'(l^{-1}(s))(l'^{-1}(s))=(\gamma'(t))t'(s)$ where we have used the fact that $t=l^{-1}(s)$. Or using Leibnitz notation: 
$\frac{d\overline \gamma}{ds}=\frac{d\gamma }{dt}\frac{dt}{ds}=\frac{\frac{d\gamma }{dt}}{\frac{ds}{dt}}$. But $\frac{ds}{dt}=\left \| \frac{d\gamma}{dt} \right \|$ from (1), so we have $$\tag 3\left \| \frac{d\overline \gamma}{ds} \right \|=1$$
