It seems like I am missing something really basic here.
I am thinking of the following 2 representations of the orientations of a 3d object (excluding reflections). Take a sphere at the origin for simplicity.
Mark 2 distinct reference points on the sphere. Represent the orientation of the sphere by the position of the first point and the direction to the second point. This gives $S^2 \times S^1$ as the space of orientations.
Start with the sphere in a reference orientation and take the rotation that puts it in the desired orientation. The space of 3d rotations is $SO(3) \cong RP^3$, a quotient of $S^3$. One representation of $SO(3)$ is axis-angle, which looks like $S^2 \times S^1$, but it is redundant. Factoring out the redundancy gives the right structure.
My problem is that I do not see a redundancy to quotient out in the first representation. If either reference point is in a different position then the whole sphere has a different orientation.
The issue in 1. is that the point on the sphere and the circle of directions aren't "topologically independent". When you choose a point on the sphere and a unit tangent vector (the "direction to the second point"), you get not $S^{2} \times S^{1}$, but the "unit tangent bundle" of the sphere, which is easily seen to be $SO(3)$. (Let $p$ denote a point of $S^{2}$ and $q$ a unit tangent vector at $p$. The cross product $p \times q$ completes an ordered orthonormal basis of $\mathbf{R}^{3}$; viewing these three vectors as columns of a matrix defines a unique element of $SO(3)$, and this map is obviously a smooth bijection with smooth inverse.)
Each of $S^{2} \times S^{1}$ and $SO(3)$ is the total space of a circle bundle over the $2$-sphere, which is why they appear superficially identical: Over an arbitrary proper open subset of $S^{2}$, the two bundles are identical.