I have found the following derivation of the Schwarzian derivatives in the book of Ovsienko and Tabachnikov:
For a diffeomorphism $\gamma$ which acts on 4 points $t_1,t_2,t_3,t_4 \in \mathbb{RP}_1$, We assume that the 4 points are spread so that $t_2,t_3,t_4$ can be defined by their distances to $t_1$ as a function of $\epsilon\in\mathbb{R}$: $t_2 = t_1 + \epsilon$, $t_3 = t_1 + 2\epsilon$ and $t_4 = t_1 + 3\epsilon$. So, the 4 points become $t_1, t_1 + \epsilon, t_1 + 2\epsilon, t_1 + 3\epsilon$ and they are related by the variable $\epsilon$. The Schwarzian derivative measures the effect of $\gamma$ on the cross-ratio as $\epsilon$ tends to zero. In other words, the Schwarzian derivative measures the cross-ratio of the points when they are infinitesimally close. To obtain the Schwarzian derivative one forms the Taylor expansion of $\Phi$ when $\epsilon$ goes to zero and keeps the first non-zero term of the expansion:
$$\Phi(\gamma(t_1),\gamma(t_2),\gamma(t_3),\gamma(t_4))= \Phi(t_1,t_2,t_3,t_4) - \epsilon^2 S[\gamma](t_1) + O(\epsilon^3) \tag 1$$
In the above equation $S[\gamma]$ is the Schwarzian derivative for $\mathbb{RP}_1$, defined by: $ S[\gamma]=\frac{\gamma'''}{\gamma'}-\frac{3}{2} \left (\frac{\gamma''}{\gamma'}\right )^2$
By construction when $\gamma$ is a linear-fractional function, $S[\gamma]$ and all the higher order terms are zero as linear-fractional exactly preserves the cross-ratio.
According to the book, the converse is also true i.e. if $S[\gamma]=0$, $\gamma$ is a linear-fractional. But I cannot see why, because in equation (1), there are higher order terms of $\epsilon$. Could anyone please tell know what I am missing here?