Let us define the differential operator $$ Z=x_1 \partial_{x_2} - x_2 \partial_{x_1}, $$ where $(x_1, x_2, x_3)$ are the standard Cartesian coordinates on $\mathbb R^3$. I would like to characterize the functions $h\in C^\infty(\mathbb S^{2})$ such that $$\tag{1} h=Zf, \qquad \text{for some }f\in C^\infty(\mathbb S^{2}).$$
The operator $Z$ is one of the generators of the rotation group $SO(3)$, in the sense that $$\tag{2} Zf(x)=\lim_{\epsilon\to 0}\frac1\epsilon \big( f(R_\epsilon x)-f(x)\big), $$ where $R_\epsilon$ is the matrix $$ R_\epsilon=\begin{bmatrix} \cos \epsilon & -\sin \epsilon & 0 \\ \sin \epsilon & \cos \epsilon & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$ The formula (2) implies that $$\int_{\mathbb S^{2}} Zf(x)\, dS(x)=0, $$ where $dS$ denotes the standard Lebesgue measure on the sphere. Thus, a necessary condition for (1) to hold is that $$\tag{3} \int_{\mathbb S^{2}} h\, dS = 0.$$ Also, since $Z$ vanishes at $(0,0, \pm 1)$, another necessary condition is $$ \tag{4} h(0,0,\pm 1)=0.$$ Are these two last conditions also sufficient?
No. A stronger condition is needed. Since the orbints of $Z$ are the circles on constant lattitude, $Zf$ must integrate to zero over each such circle. This can be seen by working in standard spherical coordinates $(\theta,\varphi):[0,2\pi)\times(0,\pi)\to S^2$. In these coordinates, $Z=\partial_\theta$, so we can apply the fundamential theorem of calculus along the lattitudes, obtaining $$ f(\theta,\varphi)=f(0,\varphi)+\int_0^\theta Zf(\theta',\varphi)d\theta' $$ Since $f(0,\varphi)=\lim_{\theta\to 2\pi^-}f(\theta,\varphi)$, we must have $\int_0^{2\pi} Zf(\theta',\varphi)d\theta'=0$. If you modify your set of conditions, you can check sufficiency by checking if the above expression gives a smooth function on $S^2$ for some choice of $f(0,\varphi)$.