Let $f:\mathbf{R}\mapsto\mathbf{P}^1$ be a function realised as $y=f(x)$, where $y$ is an affine coordinate on $\mathbf{P}^1$. Since the target space is the projective line it is desirable to lift $f$ to homogeneous coordinates, i.e. to choose $\mathbf{v}=(u(x), v(x))$ such that $u/v=f$. (Maybe it is not quite right to call this homogenous coordinates, rather this is a lifting of $f$ to the line bundle over $\mathbf{P}^1$, i.e. converting $f$ to a parameterised curve in $\mathbf{R}^2$.) Differentiating the constraint gives $$\frac{u'v-uv'}{v^2} = f',$$ which suggests taking $u'v-uv'=1$ as an additional constraint (assuming $f'>0$). This fixes $u$ and $v$ to be $$u=\frac{f}{\sqrt{f'}}$$ $$v=\frac{1}{\sqrt{f'}}$$
The added constraint says that the angular momentum of the lifted curve is constant, $\mathbf{v}'\times\mathbf{v}=1$ and this means that the acceleration vector is parallel to the position vector, $\mathbf{v''}=\lambda\mathbf{v}$. The proportionality factor $\lambda = \lambda(x)$ is $$\lambda = \frac{u''}{u}=\frac{v''}{v}=-\frac{1}{2}\frac{f'''f'-\frac{3}{2}(f'')^2}{(f')^2},$$ which is the Schwarzian with an extra factor of $-1/2$.
The proportionality factor $\lambda$ is invariant under linear transformations of $\mathbf{R}^2$. Since a projective coordinate change on the target space lifts to such a linear transformation, the Schwarzian is independent of the choice of the affine coordinate $y$ and is a projective invariant.
