There are many algorithms for computing shortest paths on polyhedral 2-manifolds.
With a student I computed the shortest paths shown below with one, the Chen-Han algorithm.
The algorithms fan out shortest paths to a frontier, mimicking the structure
(but not the details) of Dijkstra's algorithm for shortest paths in a graph. Sometimes the method is called the "continuous Dijkstra" method. These algorithms (all in $\mathbb{R}^3$) are described in many places, including
the book Geometric Folding Algorithms: Linkages, Origami, Polyhedra, Chapter 24.
I would assume the same approach will work for triangulated manifolds in arbitrary dimensions, but of course it will be significantly more complicated to implement.
I think a more fundamental way to approach the problem is by discussing geodesic curves on the surface you call home. Remember that the geodesic equation, while equivalent to the Euler-Lagrange equation, can be derived simply by considering differentials, not extremes of integrals. The geodesic equation emerges exactly by finding the acceleration, and hence force by Newton's laws, in generalized coordinates.
See the Schaum's guide Lagrangian Dynamics by Dare A. Wells Ch. 3, or Vector and Tensor Analysis by Borisenko and Tarapov problem 10 on P. 181
So, by setting the force equal to zero, one finds that the path is the solution to the geodesic equation. So, if we define a straight line to be the one that a particle takes when no forces are on it, or better yet that an object with no forces on it takes the quickest, and hence shortest route between two points, then walla, the shortest distance between two points is the geodesic; in Euclidean space, a straight line as we know it.
In fact, on P. 51 Borisenko and Tarapov show that if the force is everywhere tangent to the curve of travel, then the particle will travel in a straight line as well. Again, even if there is a force on it, as long as the force does not have a component perpendicular to the path, a particle will travel in a straight line between two points.
Also, as far as intuition goes, this is also the path of least work.
So, if you agree with the definition of a derivative in a given metric, then you can find the geodesic curves between points. If you define derivatives differently, and hence coordinate transformations differently, then it's a whole other story.
Best Answer
Not true. See: Wikipedia article: we normally define the length of a curve as a (potentially infinite) supremum of all the lengths (in the "standard" sense) of polygonal lines joining the beginning and the end of the curve and "inscribed" in the curve. We say that the curve is "rectifiable" if this supremum is finite.
With that definition, the fact that the straight line is the shortest follows pretty much immediately from the definition. Pretty much, the only slightly nontrivial bit is to prove that the length of the straight line (in this new sense) is the same as the length of the straight line (in the standard sense). This is true because, for every polygonal line "inscribed" in a straight line, the (standard) length of it is exactly the same as the (standard) length of the original straight line.
This consideration works in any $\mathbb R^n$. However, it does not scale well to other manifolds, because those may be locally homeomorphic with $\mathbb R^n$ but not locally isometric with $\mathbb R^n$. The common way of adding metric to manifolds (which is: adding a metric tensor) requires differentiability of the maps between charts in the atlas (i.e. the manifold must be differentiable), in order for the metric tensor to allow for a change in co-ordinates when switching from one chart to another. That approach scales a lot better, but requires differentiability, and in that approach, the length of the shortest path is calculated using calculus of variations. This is probably what you had in mind in your question.