The problem is three dimensional, not on a 2D sphere.
The crucial point here is that space is homogeneous. Any point in space can be taken as the origin. So suppose you have an observer. You are perfectly allowed to decide that the position of this observer is the origin of your system of coordinated. So he is sitting at the point $r_0=0$ and there, $\phi$ and $\theta$ are not defined.
Now what is a geodesic passing through this point ? Well, they are all the "straight lines" where $r$ varies from $0$ to positive infinity, with a fixed value of $\theta$ and $\phi$. This is how a photon emitted from the observer will propagate. Why would it not ? Space is homogeneous. Of course this is just a half-line, the full line will consist of the two half lines $(\theta, \phi)$ and $(-\theta, \phi+\pi)$. These are not all the geodesics, just those geodesics that pass through this observer.
The fact that space is curved, in this case, does not alter the "straight line propagation", because everything is homogeneous. "Curved space" means that if you measure the length of a circle, you will not find its diameter times $\pi$. In the coordinates of a different observer, the "straight lines" of constant $\theta$ and $\phi$ of the first observer do not look "straight". But geodesics passing through himself will be "his straight lines", constant $(\theta, \phi)$ in his coordinate system.
Let me be more precise. Suppose you consider the geodesic 𝜃=𝜋/2 (or 𝜃=−𝜋/2). For these values of 𝜃, 𝜙 is not defined. Now this is the trajectory of a photon going straight up (or down). Now in the real Universe there will be a star here, a galaxy there, the photon will deviate. But in the approximation of an homogeneous Universe, all directions are equivalent. To deviate from 𝜃=𝜋/2 (or −𝜋/2) the photon would have to choose some value for 𝜙. But which one? They are all equivalent ! Since it cannot choose a 𝜙 it must stay on 𝜃=𝜋/2.
And since you can always choose your system of coordinates so that the direction you consider is (𝜃=𝜋/2, 𝜙 undefined), then the geodesics starting in this direction can never "choose" a 𝜙 to deviate in that direction rather than any other. Remember all this argument depends on the assumption that the universe is homogeneous and isotropic, that is all directions are identical. If the universe is homogeneous but not isotropic one could not choose arbitrarily any direction to be 𝜃=𝜋/2. So in fact it only holds for homogeneous and isotropic Universes.
However, when one speaks of homogeneous Universes ,it is usually understood that they are isotropic too. Homogneous Universes which are not isotropic can be described mathematically, but they are weird. In such Universes, indeed, geodesics going through an observer might well not be with fixed $\theta$ and $\phi$. But clearly the intent of Kolb and Turner were an homogeneous and isotropic Universe even if they did not point it out.
Well, the FLRW metric is definitely both homogeneous and isotropic so my argument does indeed hold.