So, your first step is indeed correct, but the one you are not sure about is wrong. You are probably confused by your own notation, I prefer to write $\partial_i=\frac{\partial}{\partial x_i}$ instead of $\partial x_i$. And remember that repeated indices are being summed over, so e.g. $u_ju_j=\mathbf{u}\cdot\mathbf{u}$. Given that, you have
\begin{equation}
u_j\partial _i u_j-u_j\partial_j u_i
=(1/2)\partial_i(u_j u_j)-(\mathbf{u}\cdot \nabla)u_i
=(1/2) \partial_i (\mathbf{u}\cdot\mathbf{u})-(\mathbf{u}\cdot \nabla)u_i,
\end{equation}
which (taking into account your earlier work) is the $i$-th component of the equation
\begin{equation}
\mathbf{u}\times(\nabla\times \mathbf{u})=\frac{1}{2}\nabla(\mathbf{u}\cdot\mathbf{u})-(\mathbf{u}\cdot \nabla)\mathbf{u}
\end{equation}
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.
Best Answer
You've got up to $$(\nabla\times A)\cdot(\nabla\times A)=\left(\frac{\partial A_j}{\partial x_m}\right)\left(\frac{\partial A_j}{\partial x_m}\right)-\left(\frac{\partial A_j}{\partial x_m}\right)\left(\frac{\partial A_m}{\partial x_j}\right).$$ Note that this is a double sum over $j$ and $m$ (this is why I replaced your $k$ by a $j$, to make the Einstein summation notation correct).
The Jacobian $J$ has components $J_{ij}=\frac{\partial A_i}{\partial x_j}$. So we may write $$(\nabla\times A)\cdot(\nabla\times A)=J_{jm}J_{jm}-J_{jm}J_{mj}.$$ Now $J_{jm}J_{jm}$ is the sum of the diagonal elements of the matrix $JJ^\top$, i.e. $\operatorname{Tr}(JJ^\top)$ and likewise $J_{jm}J_{mj}=\operatorname{Tr}(J^2)$.
So what we have is $\operatorname{Tr}(JJ^\top)-\operatorname{Tr}(J^2)$.