You want a normal vector rather than "tangent vector." You make an $(n-1)$ by $n$ matrix with row $i$ given by $x_i - x_n.$ Then the normal vector is the null space of this matrix. The fast way to do this is Gauss elimination, but it can also be done as a recipe by pretending the matrix is square, and find the entries of the proper column of the adjoint matrix, given by cofactors with certain $\pm$ signs. The traditional cross product is often taught this way in physics classes.
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
It sounds like you're just trying to carry out the picture of spherical geometry in 3-D to 4-D.
In the 3-D picture, the surface of the unit sphere is taken to be the set of points, and the "lines" are the great circles. Any two points which aren't antipodal determine a unique great circle.
The way to look at the great circle in that case is as the intersection of a plane through the origin with the sphere. This is suggestive then that you are looking for the intersection of the plane through your two points and the origin in 4-D, and the line should be the intersection of this plane with the sphere $S_3$.
Looking at it from this perspective, you can believe that two nonantipodal points with the origin of 4-space form a noncollinear set of three points, and hence a unique plane.