I am not sure if this helps, but I just found that you can actually use the contracted epsilon identity to get a pretty good vector/dyadic representation of the chain rule. The nice thing about this particular identity (at least to me) is that it does what any good chain rule should do, and applies the operator in question to the argument.
Please allow me to preempt this by saying that I am not that familiar with the conventions of dyadic notation, but I will present what I have figured out in index notation form, so that if anyone wants to go in, and fix my notation, they will know how to. Anyway, here is what I found:
\begin{equation}
\nabla \times \left(\mathbf{A} \circ \mathbf{B}\right) = -\left(\nabla_\mathbf{B}\cdot \mathbf{A}\right)\left(\nabla\times \mathbf{B}\right) + \left(\frac{\partial \mathbf{A}}{\partial \mathbf{B}}\right)^T\left(\nabla \times \mathbf{B}\right) + \left(\frac{\partial \mathbf{A}}{\partial \mathbf{B}}\right)^T \!\!\!\begin{array}{c}
_\cdot \\
^\times\end{array}\!\!\!\left(\nabla \mathbf{B}^T\right)
\end{equation}
The vertical operator notation means that I am crossing the second index of the left tensor with the 1st index of the right tensor, and then contracting the 1st index of the left tensor with the 2nd index of the right tensor.
Working it out in index notation, we start with
\begin{equation}
\epsilon_{ijk}\frac{\partial A_j}{\partial B_l}\frac{\partial B_l}{\partial X_k}
\end{equation}
What we want is to have the levi-civita symbol apply to B. The first step for this is to "free" the l and k indices. We can do this, using the kronecker delta
\begin{equation}
\epsilon_{ijk}\frac{\partial A_j}{\partial B_l}\frac{\partial B_l}{\partial X_k} = \delta_{k_0k_1}\delta_{l_0l_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}}
\end{equation}
Now, we can get to the contracted epsilon identity by adding appropriate cancelling terms to the RHS:
\begin{equation}
\left(\delta_{k_0k_1}\delta_{l_0l_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}} - \delta_{k_0l_1}\delta_{l_0k_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}}\right) + \delta_{k_0l_1}\delta_{l_0k_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}}
\end{equation}
We can then use the contracted epsilon identity to rewrite this as
\begin{equation}
\epsilon_{rk_0l_0}\epsilon_{rk_1l_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}} + \delta_{k_0l_1}\delta_{l_0k_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}}
\end{equation}
Now, we are halfway there, for $\epsilon_{rk_1l_1}$ applies to $\frac{\partial B_{l_1}}{\partial X_{k_1}}$
Notice also, that when we juxtapose $\epsilon_{rk_0l_0}$ and $\epsilon_{ijk_0}$ we can apply the contracted epsilon identity once again. Combining these two facts gives us
\begin{equation}
\left(-\delta_{ri}\delta_{l_0j} + \delta_{rj}\delta_{l_0i}\right)\frac{\partial A_j}{\partial B_{l_0}}\epsilon_{rk_1l_1}\frac{\partial B_{l_1}}{\partial X_{k_1}} + \delta_{k_0l_1}\delta_{l_0k_1}\epsilon_{ijk_0}\frac{\partial A_j}{\partial B_{l_0}}\frac{\partial B_{l_1}}{\partial X_{k_1}}
\end{equation}
The expression in vector/dyadic notation above follows directly from this.
The hypotheses of the Divergence Theorem require that the vector field be defined and $C^1$ on all of the solid $V$, but here the vector field ${\bf A} := \frac{1}{r^2} {\bf a}_r$ is not defined at zero (indeed, as $r \to 0$, $\bf A$ is unbounded). Thus, the Divergence Theorem does not apply. The computation $$\iint_S {\bf A} \cdot d{\bf S} = 4 \pi$$
is, by the way, correct (for the usual outward-pointing orientation of the sphere $S$).
There's an analogous example for Green's Theorem: The vector field $${\bf X} := \bigg(\underbrace{-\frac{y}{{x^2 + y^2}}}_P \partial_x + \underbrace{\frac{x}{{x^2 + y^2}}}_Q \partial_y\bigg)$$ satisfies $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$, but the contour integral along the unit circle (oriented anticlockwise) is $$\oint_{\Bbb S^1} {\bf X} \cdot d{\bf s} = 2 \pi,$$ which does not coincide with $$\iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = 0,$$
where $D$ represents the unit disk. Again, there is no contradiction, because $\bf X$ is badly behaved near zero and so does not satisfy the hypotheses of Green's Theorem.
Best Answer
It would be more precise to say: the derivative of the divergence of a vector field with respect to that field. (You say "function" but then $F$ turns out to be a vector field.) In symbols, it could be expressed as $D_F (\nabla \cdot F)$ although notation varies. It helps to write fields in derivative notation, $$F=\sum_{i=1}^3 f_i \frac{\partial}{\partial x_i} \tag1$$ where $f_1,f_2,f_3$ are scalar functions, the components of the field. Then the divergence of $F$ is expressed by $$\nabla \cdot F=\sum_{i=1}^3 \frac{\partial f_i }{\partial x_i} \tag2$$ The derivative of (2) with respect to the vector field (1) comes out as the double sum $$D_F(\nabla \cdot F)=\sum_{i=1}^3 f_i \frac{\partial}{\partial x_i} \sum_{j=1}^3 \frac{\partial f_j }{\partial x_j} = \sum_{i,j=1}^3 f_i \frac{\partial^2 f_j }{\partial x_i\partial x_j} \tag3$$ The expression (3) is not zero in general. For example, take $f_i=x_i^2$ for $i=1,2,3$: then the sum (3) evaluates to $2(x_1^2+x_2^2+x_3^2)$.