Instead of a 4D hyperbolic geometry (saddle-shaped), is it conceivable a spacetime with negative curvature has a 4D spherical geometry where the perspective is from the "inside" of the sphere rather than the "outside"?
The curvature of a surface is defined as the product of its principal curvatures. Whether they are both positive (on the outside if a sphere) or both negative (on the inside), the product will still be positive.
The curvature is intrinsic to the surface, as @Cornifold mentions above. It doesn't matter from which side you see it. This is a fundamental mathematical theorem, look e.g. on its Wikipedia page
As you point out in your own comment, the distance and angular measures on a sphere are the same whether you look at it from the inside or the outside. This is basically a different way of saying the same thing - that the curvature is unchanged. This is, in some sense, exactly what curvature means.
Your question should dissolve away when you understand the meaning of curvature.
When we talk about spheres and saddles, we imagine them being in three-dimensional space. We can use a Cartesian 3D system and describe the surfaces completely by some function $F(x,y,z) = 0$. For example, a unit sphere would be described by $$ x^2 + y^2 + z^2 - 1 = 0$$
Mathematically, what we are doing is called embedding. We embedded a two dimensional surface in three dimensional space. It's a mathematical fact that we can embed such surfaces, i.e. manifolds in higher-dimensional ordinary Euclidean spaces.
But here's the thing. We don't have to think about surfaces as being embedded in some higher-dimensional space. The sphere, for example, as a two-dimensional manifold is what it is whether you embed it or not, it's the surface that matters. The curvature of a manifold does not depend on its embedding, it is an intrinsic property of the manifold itself.
So, you see, there is no outside and inside, not really. One way to measure curvature of the sphere would be to draw a circle on it and compare its circumference to its area. The ratio would, of course, be smaller than on a sheet of paper, which is flat space. Similarly, in three-dimensional space, you would compare the volume within a sphere to its surface area.
Therefore, when talking about curvature of 4D spacetime, don't imagine it as being a hypersurface in a higher-dimensional space, it's not necessary.
On a sphere, if you take it as a 2D representation of curved 3D space, it is possible to move straight forward and end up where you started from (which would also be the case in a nonexpanding closed universe; in reality the universe expands too fast for even a light ray to catch up with the increasing curvature circumference) but in an open universe you can't, so the sphere would be the wrong representation for an open universe, but the right one for a closed.