Right cone, you are at A and need to complete a revolution before reaching the bottom B. What is shortest distance AB? You are on a mountain that is a right cone shape. You are trying to get to B and you are somewhere up the mountain A such that you lie on the line OB.
The line AB must do one full revolution of the mountain and is the shortest distance.
What is the general formula for finding any AB distance? How much of that distance is uphill/downhill?


When uncurled the cone looks like the shape below with A and B demonstrated. With A being a distance $x$ along the line OB. I appreciate that the straight line is the shortest distance but how do I work that distance out? 
I am not sure where to begin with uphill/downhill part.


Henry has answered most of your questions, except how to calculate the distance of the path. To find the distance you need to find the angle $\theta$ of the circular sector. 

We know the circular arc of the sector has the same length as the circumference of the cone's base circle, so $\theta s = 2\pi r$ giving: $$\theta = \frac{2 \pi r}{s}$$
If the point $A$ is located a distance $d$ from the cone vertex, we can now use the Law of Cosines to find the path distance x: $$x^2 = s^2 + d^2 - 2sd \cos(\frac{2 \pi r}{s})$$
EDIT
To find the uphill distance $x_u$ we use the method given by Henry, i.e. drop a perpendicular from $O$ to $AB$, meeting at $C$. The height $h$ of $OC$ can be found by the area formulas for a triangle: $$\text{Area} = \frac{1}{2}\cdot s \cdot d \cdot \sin(\theta) = \frac{1}{2}\cdot x \cdot h$$
giving $$h = \frac{s \cdot d \cdot \sin(\theta)}{x}$$
We then see that $x_u^2+h^2=d^2$ or $$x_u = \sqrt{d^2-h^2}$$
