A surface is said to be uniformly negatively curved if the angles of a triangle ($\alpha, \beta, \gamma$) satisfy the relation
$\alpha +\beta + \gamma = \pi - A/R^2$
Where $A$ is the area of the triangle and $R$ is the radius of curvature.
Why does the interior surface is a sphere not satisfy this? In other words, how can the sum of the angles be invariant on the triangle's position on the surface, if $R$ is not constant? I assume $R$ is not constant for uniform negative curvature, because if it were, wouldn't we be on the interior of a sphere?
The interior of the sphere does satisfy this, but the interior has positive curvature. $R$ is the radius of the sphere, but $1/R^2$ is the Gaussian curvature.
Gaussian curvature is an intrinsic concept, so you get the same thing no matter whether you draw on the inside or the outside of a sphere. If you define it extrinsically, you do so as the product of the curvature of the two circles osculating with the curvature lines. For the outside of the sphere, you get one pair of radii, for the inside presumably the other pair, but that doesn't change the product at all. Thus the curvature is the same.
This is also related to the fact that you have a $R^2$ in that formula. In some interpretations, switching between inside and outside might correspond to switching the sign of the radius, but that does not affect the formula. To obtain negative curvature you need an imaginary radius. In fact you can get quite some distance by assuming a sphere of imaginary radius, with some tweaks to switch between real and imaginary numbers as appropriate. The result is the hyperboloid model of hyperbolic geometry.