Let $\mathbb{S}^{2}$ be $$\mathbb{S}^{2}:=\left\{\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)\in \mathbb{R}^3\middle| x^2+y^2+z^2=1\right\}$$ ,and let $\mathbb{H}$ be $$\mathbb{H}:=\left\{\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)\in \mathbb{R}^3\middle| x^2+y^2-z^2=-1 ,z > 0 \right\}$$
Here, the $\mathbb{R}^3$ be a three dimensional Euclidean space, and the $\textbf{O}$ means the origin of $\mathbb{R}^3$.
Then, according to my text book , following very beautiful theorems are satisfied on both $\mathbb{S}^2$ and $\mathbb{H}$. On the other hand, there are various curved surfaces other than the spherical surface and the (upper side of) biplane hyperboloid; for example, one-leaf hyperboloid, paraboloid, torus, projective plane, Mobius ring, etc.
【My Question】
(1) What are some other surfaces other than $\mathbb{S}^2$ and $\mathbb{H}$ that have the following theorem?
(2) In particular, is it possible to "define" a straight line in other surfaces in a way similar to the following theorem? For example, one-leaf hyperboloid, paraboloid, torus, projective plane, Mobius ring.
(3) If possible, How is that?
【Theorem 1】
The straight line on $\mathbb{S}^{2}$ shall be the intersection of "$\mathbb{S}^{2}$" and "the plane through the $\textbf{O}$".
The intersection of "$\mathbb{S}^{2}$" and "the plane through the $\textbf{O}$" shall be the straight line on $\mathbb{S}^{2}$.
【Theorem 2】
The straight line on $\mathbb{H}$ shall be the intersection of "$\mathbb{H}$" and "the plane through the $\textbf{O}$".
The intersection of "$\mathbb{H}$" and "the plane through the $\textbf{O}$" shall be the straight line on $\mathbb{H}$.(See the following Fig.1.)
Here, the planes on Th1 and Th2 are the plane in the sense of Euclidean space.
Fig.1 is quoted from this site with some modifications, representing the relationship between "the line on $\mathbb{H}$" and "the plane through the $\textbf{O}$.