Oh, right, since we are lifting maps $S^2\to \mathbb{R}P^2$. So the ratio of these two lifts is the antipodal map. Hence only one of them preserves the orientation. For example, for the fixed Riemannian metric there is a canonical choice.

Aren't diffeomorhisms of the disks are supposed to agree on the circle? Do we account for it here?

I just realized that my comment is incorrect as stated but you were faster:-)

Yes, that makes perfect sense! Thanks! So for the sphere the same construction gives us the bundle $\text{Diff}(D^2)\times \text{Diff}(D^2) \to \text{Diff}(S^2) \to \text{Emb}(S^1, S^2)$?

I am just curious, what is the reason they add $\text{mod }F$ there? For example, in the case $A=1$, $B=0$ I would expect the solution to satisfy $F_x =0$ but instead we get $F_x$ is proportional to $F.$

@Robert Bryant: Yeah I've understood your solution and it is really nice approach. Also it turned out that in my research the function g is not an arbitrary flat function but a function of the form $g = \alpha(pq),$ where $\alpha$ is a flat function of one variable. In this case everything is a little bit simpler, for example, the function $g$ is even for any $\alpha.$ But still, it is quite interesting that this statement is true for an arbitrary flat function.

@IgorKhavkine, yeah, that's right.