For your characterisation of convex functions, you are using the fact that the straight line is an optimiser of the inequality. More precisely, we try to solve
$ |xz| f(y) = |xy| f(z) + |yz|f(x) $
where $x,y,z$ are collinear. Without loss of generality, we can assume that $x,y,z\in \mathbb{R}$. Write $z = y + \delta y$ and $x = y - \delta y$ and doing a second order Taylor expansion we get that
$ f'' = 0 $
and so a necessary condition is that $f$ is linear. (We are making assumption of differentiability etc.) We then check that all linear functions $f$ satisfy this condition. And we can use it as an upper envelope of "convexity" between two points.
Now let us try to do the same with spherical convexity. We have $ \sin( 2 \delta y) f(y) = \sin (\delta y) f(y + \delta y) + \sin (\delta y) f(y - \delta y) $ Taking the Taylor expansion to $O(\delta y^3)$ on both sides we get $ \left[ 2 \delta y - \frac{8}{6} (\delta y)^3\right] f(y) =_{O(\delta y^3)} \left[ \delta y - \frac{1}{6} (\delta y)^3\right] \left[ 2 f(y) + f''(y) (\delta y)^2\right] $ which simplifies to $ -f(y) = f''(y) $ or that $f(y) = A \sin (y + B)$. We check that these functions indeed verify the hypothesis: assuming $z \geq y \geq x$, and $A = 1$ since the expression is scale invariant,
$ \sin( z - x) \sin (y + B) \overset{?}{=} \sin( y - x) \sin (z + B) + \sin(z - y) \sin(x + B) $ or $ \left( \sin z \cos x - \cos z \sin x \right) \left(\sin y \cos B + \cos y \sin B\right) \overset{?}{=} \left( \sin y \cos x - \cos y \sin x\right) \left( \sin z \cos B + \cos z \sin B\right) + \left( \sin z \cos y - \cos z \sin y\right) \left(\sin x \cos B + \cos x \sin B\right) $ which one can simply check to hold.
Therefore, the interpretation of "spherical convex" that is analogous to the characterisation of convex function is exactly like the standard convex case, except where the comparison is made with the straightline through $(x,f(x))$ and $(z,f(z))$, we compare against the unique function $g(s) = A \sin(s + B)$ defined over the geodesic connection $x$ and $z$, with $s$ an arclength parametrisation, such that $g(s(x)) = f(x)$ and $g(s(z)) = f(z)$.