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.
Sorry I cant comment, because my reputation is not big enough, so I will just answer...
Imagine you throw a ball inside the vector field. This ball starts to move alonge the vectors and the curl of a vectorfield is a measure of how much the ball is rotating.
The curl gives you the axis around which the ball rotates, its direction gives you the direction of the orientation (clockwise/counterclockwise) and its length the speed of the rotation.
This is some curde model to give you some imagination about the curl...
Best Answer
I thought it might be useful to present way forward that I prefer. Here, we begin by introducing notation.
Let $(x_1,x_2,x_3)$ be Cartesian coordinates. We designate by $\hat x_i$, a unit vector along the $x_i$ axis and by $\partial_i$ the partial derivative with respect to $x_i$.
Then, using the convention of summing over repeated indices, the $i$'th component of the vector $\nabla \times (\vec B\times \vec C)$ can be written as
$$\begin{align} \hat x_i\cdot \left(\nabla \times (\vec B\times \vec C)\right)&=\hat x_i\cdot \left(\hat x_j\times (\hat x_k\times \hat x_\ell)\right)\partial_j(B_kC_\ell)\tag 1\\\\ &=\left(\delta_{ik}\delta_{j\ell}-\delta_{i\ell}\delta_{jk}\right)\left(B_k \partial_j(C_\ell)+C_\ell \partial_j(B_k)\right)\tag 2\\\\ &=B_i\partial_j(C_j)-B_j\partial_j(C_i)+C_j\partial_j(B_i)-C_i\partial_j(Bj)\tag 3\\\\ &=B_i(\nabla \cdot \vec C)-(\vec B\cdot \nabla)C_i+(\vec C\cdot \nabla)B_i-C_i(\nabla \cdot \vec B) \tag 4 \end{align}$$
In going from $(1)$ to $(2)$, we used the vector triple product rule (presumed established) while in going from $(2)$ to $(3)$ we used the sifting property of the Kronecker Delta.
Since $(4)$ is true for all $i$, then upon multiplying by $\hat x_i$ and summing over $i$, we find the coveted identity
$$\nabla \times (\vec B\times \vec C)=\vec B(\nabla \cdot \vec C)-(\vec B\cdot \nabla)\vec C+(\vec C\cdot \nabla)\vec B-\vec C(\nabla \cdot \vec B)$$