The SL2 flag charts realize the stated rho shift

The weight of the unique positive root is $2\rho$, so $\rho (h)=1$. With $d=-c-1$, the associated line is ${\mathcal{O}}_X((c+1)\rho )={\mathcal{O}}_{{\mathbb{P}}^1}(d)$ by [F2]. On $U_0$, the inverse upper-unipotent matrix sends $(1,z)$ to $(1-tz,z)$; therefore its action on a local coefficient $g(z)e_0$ is $(1-tz{)}^{d}g(z\mathrm{/}(1-tz))e_0$. Differentiating at $t=0$ gives $z^2{\mathrm{\partial }}_z-dz=z^2{\mathrm{\partial }}_z+(c+1)z$ for $e$. The inverse lower unipotent sends $(1,z)$ to $(1,z-t)$, giving $-{\mathrm{\partial }}_z$ for $f$. The inverse diagonal matrix sends $(1,z)$ to $(e^{-t},e^{t}z)$; the homogeneous frame contributes $e^{-dt}$, so differentiation gives $2z{\mathrm{\partial }}_z-d=2z{\mathrm{\partial }}_z+(c+1)$ for $h$.

On $U_0\cap U_1$, the frames satisfy $e_1=z^d{e}_0$, so a coefficient $g_1(w)$ on $U_1$ corresponds to $g_0(z)=z^d{g}_1(1\mathrm{/}z)$ on $U_0$. If $T{g}_1=z^d{g}_1(1\mathrm{/}z)$, direct differentiation gives $T^{-1}{\mathrm{\partial }}_{z}T=dw-w^2{\mathrm{\partial }}_w\mathrm{.}$ Conjugating the three operators from step 1.1 by $T$ and using $d=-(c+1)$ gives, respectively, $-{\mathrm{\partial }}_w$, $w^2{\mathrm{\partial }}_w+(c+1)w$ and $-2w{\mathrm{\partial }}_w-(c+1)$. Thus the two formulas glue. Since they are obtained by differentiating the algebraic $G$-action, their commutators are the brackets in [F1]; direct substitution gives the same relations.

At $c=-1$ one has $d=0$, so the line bundle is the trivial bundle and [F2] gives ${\mathcal{D}}_{-\rho }=\mathrm{Diff}({\mathcal{O}}_X)$, as claimed.