Let $\mathfrak{X}_{CK}^{\perp}$ be the space of vector fields on $S^2$ that are $L^2$-orthogonal to conformal Killing vector fields. Let $\mathfrak{X}_{CK}$ be the 6-dimensional space of conformal Killing vector fields on $S^2$.
Can we find a vector field $Y \in \mathfrak{X}_{CK}^{\perp}$ and a vector field $W \in \mathfrak{X}_{CK}$ that is not Killing such that
$$\mathrm{div}(Y) = \mathrm{div}(W)$$
I think that this is not possible:
Per my comment on Divergence of conformal Killing vector fields on $S^2$ and the spherical harmonics you want to solve $$ \textrm{div} (Y) = -2a\cdot x $$ for $Y$ orthogonal to conformal-KVF's and $a \in \mathbb{R}^3$ fixed (nonzero). Suppose you can do this. Then, we find that $$ Y = a^T + W $$ for $W$ divergence free. From this question https://math.stackexchange.com/questions/1898371/divergence-free-vector-field-on-a-2-sphere you can see that $W = J \nabla f$ for some function $f \in C^\infty(S^2)$ (for $J$ the complex structure on $S^2$). Now, by assumption, $Y$ is orthogonal to $a^T$: \begin{align*} 0 & = \int_{S^2} Y\cdot a^T \\ & = \int_{S^2}|a^T|^2 + (J \nabla f) \cdot a^T \\ & = \int_{S^2}|a^T|^2 - \nabla f \cdot J a^T \\ & = \int_{S^2}|a^T|^2 + f \textrm{div}(J a^T). \end{align*} However, note that $Ja^T$ is a KVF and thus divergence free. This implies that $\int_{S^2}|a^T|^2 = 0$. This is a contradiction.
