I'm trying to figure out if there is a family of map projections which preserve local distances: in other words a family of functions $f \in S^2 \rightarrow K, K \subseteq \mathbb R^2$ such that for any tangent vector $\vec v$ at a point $p$ in $S^2$, the norm of $f_p[\vec v]$ is the same as the norm of $\vec v$.
Equivalently I suppose you can do it in spherical and Cartesian coordinates and demand $x_\theta x_\theta ~ + y_\theta y_\theta = 1$, $x_\phi x_\phi ~ + y_\phi y_\phi = \sin^2\theta$, and $x_\phi x_\theta + y_\phi y_\theta = 0$.
I know that at least one set of such maps exists: you can stereographically project $S^2 \rightarrow \mathbb R^2$ and then radially distort the map in polar coordinates to make the $\theta$-components of tangent vectors equal to the radial components. Is there a name for the resulting map projections?
Is there a subfamily of the broader set which is conformal? I know you can't get both conformal and equal-area but is there a similar result for conformal and locally-distance-preserving?
Turning my comments into an answer.
If you preserve local distances, then you preserve path lengths as these are integrals over line elements. Which means you map geodesics to geodesics, because if you did not, then you could find a shorter path in one world which would map to a path that is shorter than a geodesic in the other. This obviously can't be, so by contradiction you map geodesics to geodesics.
Since there is no projection which maps all great circle arcs on $S^2$ to line segments on $\mathbb R^2$, your map projection can't exist. I can't see why you think that what you describe as an example should be possible. How would you distort $\theta$ components without affecting $\phi$ components?
See also the Wikipedia article on map projections which indicates that probably the most you can hope for is equal distance to two arbitrarily chosen points. So speaking in terms of probability measures, a randomly chosen pair of points will almost never (i.e. with probability zero) have its distance preserved.
Or you allow local isotropic scaling, in which case you'd be back to all conformal maps. Then you have your stereographic transformation, which strictly speaking maps the sphere to the one-point compactification of the plane, or equivalently maps the punctured sphere to all of $\mathbb R^2$.