How much of the Earth's atmosphere is visible from its surface If I observe a meteor streak across the sky, how can I estimate the length of its trail in kilometres, given say its approximate height of 15km from the surface, and its apparent length of a quarter of the night sky.
 A: 
How much of the Earth's atmosphere is visible from its surface

All of it.
Though you need either a lot of observers or a lot of time.
I assume you mean without taking into account clouds, haze and other atmospheric issues. Or not, depending on what wavelengths are "visible" to our observer(s). It kinda depends on whether the observer is a cat or a squid.
If we restrict ourselves to a single stationary observer - it depends how tall they are and where exactly they are standing.

Image by liftarn/Acdx
If the observer is 5' 7" (170 cm) and standing on a barely awash shoal in the middle of the pacific, the horizon is about 4.7 km away.
Calculating the volume of atmosphere visible to the observer is merely a (somewhat boring) calculation. You should probably correct for the fact that the Earth is not a sphere. You'd have to make some arbitrary definition of how far the atmosphere extends upwards.
Note that because the atmosphere is three dimensional, you can't easily apply 2D geometry to questions of this sort ("how much of").

If I observe a meteor streak across the sky, how can I estimate the length of its trail in kilometres

Not easily. You need several observations from differing vantage points.

Using the footage and the location of an impact into Lake Chebarkul, Jorge Zuluaga and Ignacio Ferrin, from the University of Antioquia in Medellin were able to use simple trigonometry to calculate the height, speed and position of the rock as it fell to Earth.

- BBC: Chelyabinsk meteor

Using evidence gathered by one camera at the Revolution Square in the city of Chelyabinsk and other videos recorded by witnesses in the close city of Korkino, we calculate the trajectory of the body in the atmosphere and use it to reconstruct the orbit in space of the meteoroid previous to the violent encounter with our planet

- A preliminary reconstruction of the orbit of the Chelyabinsk Meteoroid

given say its approximate height of 15km from the surface

That is problematic because meteors are not in a circular orbit (or trajectory) and therefore do not stay a constant height above the surface as they travel across a quarter of the visible sky.
If you have some means of measuring height from surface, you can probably use that same means to measure distance from observer at each point in its visible journey. Together with an angular measurement that would allow you to calculate the distance traversed.

Other refs
RECONSTRUCTING THE CHELYABINSK METEOR’S PATH, WITH GOOGLE EARTH, YOUTUBE AND HIGH-SCHOOL MATH
A: If you know the distance (15km) the approximate computation is easy with a bit of trigonometry. A quarter of the night sky is a bit imprecise: apparent sizes are measured in angle units, that is, degrees or radians.
The sky (night or day) is a maximum of 180º, so a quarter of that would be more or less 45º. Now, assume the following:


*

*the trail is a straight line.

*the center of the trail is at 15km of the observer.

*the trail is perpendicular to the line segment from the observer to the center of the trail.


Thus, you can easily measure half a trail, as the tangent of half the angle times the distance:
$ L = 2\ D\ tan(\alpha/2 ) $
being $\alpha = 45º, D=15\ km$ 
That would be about $12.4\ km$
If you don't mind the extra error, you can just use the simpler:
$ L = D\ tan(\alpha) $
In your example that would be just $15\ km$.
