Geometric interpretation of $\operatorname{SO}(3)/\operatorname{SO}(2)=S^2$ I understand (in a hand-waving sense) the argument that if an element of $\operatorname{SO}(3)$ is a general rotation in 3D characterized by a direction in 3D and a rotation angle, then "dividing out" the degree of freedom associated with the rotation ($\operatorname{SO}(2)$) leaves a direction in space, and the set of all possible directions can be mapped to the surface of a sphere.
However, can I understand this statement by choosing a particular set of rotations about the $z$-axis (which is $\operatorname{SO}(2)$) as my subgroup of $\operatorname{SO}(3),$ and then forming all cosets of it ? Each coset would be a specific rotation in 3D applied to the set of all rotations about the $z$-axis. I think that I should be able to see a geometrical way that each coset can be mapped to one (and only one) point on the surface of the sphere ($S^2$). However, I can't quite make a geometrical connection.

The group $SO(3)$ naturally acts on $3$ dimensional space, and preserves the unit sphere within this. Furthermore, it acts transitively on this unit sphere, so picking a point $x$ on $S^2$, mapping $g\in SO(3)$ to $g.x$, we have a continuous surjective map $SO(3)\rightarrow S^2$. One may check that the fibre over a point is a copy of the circle group $SO(2)$, a conjugate of the stabiliser of $x$ (or a coset of the stabiliser, if you prefer), so in this sense, each point on $S^2$ gets a copy of $SO(2)$, and together these fill out all of $SO(3)$.
Morally, this is just the orbit stabiliser theorem of group theory, after noticing that the proof holds in the category of topological spaces, if your group action is nice.
We can also notice that our fibres seem to change and depend on our point $x$ chosen, and this reflects that although this map is locally a product, it is not globally so. That is, as topological spaces, we do not have $SO(3)\cong SO(2)\times S^2$, even though we can find isomorphisms when we restrict to the fibres over each hemisphere of $S^2$.
