How to simplify expressions with del (or nabla) in them?

I'll expand what I've commented above here.

Using suffix notation and summation convention (since we are working with $mathbb{R}^n$ there is no need to distinguish upstairs and downstairs indices, so just write everything downstairs), you can get, for example, begin{align*}require{color} [nablatimes(mathbf{A}timesmathbf{B})]_i &=epsilon_{ijk}partial_j(mathbf{A}timesmathbf{B})_k\ &=epsilon_{ijk}epsilon_{kell m}partial_j A_ell B_m\ &=(delta_{iell}delta_{jm}-delta_{im}delta_{jell})partial_j A_ell B_m\ &={color{red}partial_j A_i B_j}-{color{blue}partial_j A_j B_i}\ end{align*} Note that, by convention, $partial$ acts on everything to its right. Whereas in the case of $mathbf{C}times(mathbf{A}timesmathbf{B})$ we would just "take out" the $A_i$ and $B_i$ from the red and blue term respectively, here we can't move them in front of the differential operator without paying for the noncommutativity: begin{align*} {color{brown}partial_j A_i} B_j&={color{brown}A_ipartial_j}B_j+({color{brown}partial_jA_i})B_j\ &=A_ipartial_jB_j+B_jpartial_jA_i\ partial_j A_j B_i&=B_ipartial_jA_j+A_jpartial_jB_i. end{align*} So $$ nablatimes(mathbf{A}timesmathbf{B})=mathbf{A}(nablacdotmathbf{B})-mathbf{B}(nablacdotmathbf{A})+(mathbf{B}cdotnabla)mathbf{A}-(mathbf{A}cdotnabla)mathbf{B} $$ and you can see the additional terms are precisely what we get from moving something behind a $nabla$ to in front of that $nabla$.

Now we have done the calculation, you might reasonably ask the question: Can one immediately get from $$ mathbf{C}times(mathbf{A}timesmathbf{B})=mathbf{A}(mathbf{C}cdotmathbf{B})-mathbf{B}(mathbf{C}cdotmathbf{A}) $$ to a formula for $nablatimes(mathbf{A}timesmathbf{B})$? To start with, we check the naive substitution still gives terms that make sense (i.e., it doesn't leave dangling $nabla$). Then we see the formula involves pushing $mathbf{A}$ (or $mathbf{B}$ in the second term) in front of $mathbf{C}$, so we need to compensate that by having something from $nablamathbf{A}$ (or $nablamathbf{B}$). So the formula has to read something like $$ mathbf{A}(nablacdotmathbf{B})-mathbf{B}(nablacdotmathbf{A})+(nablamathbf{A})astmathbf{B}-(nablamathbf{B})astmathbf{A} $$ where $ast$ does some contraction between the rank-2 tensor and the vector. Now it is not hard to see in $(nablamathbf{A})astmathbf{B}$ the $mathbf{B}$ has to contract with the $nabla$ rather than $mathbf{A}$ (because the term we are correcting has that), hence we obtain $$ mathbf{A}(nablacdotmathbf{B})-mathbf{B}(nablacdotmathbf{A})+(mathbf{B}cdotnabla)mathbf{A}-(mathbf{A}cdotnabla)mathbf{B}. $$

Curl of a curl: Similarly, $$ mathbf{C}times(mathbf{A}timesmathbf{B})=mathbf{A}(mathbf{C}cdotmathbf{B})-(mathbf{C}cdotmathbf{A})mathbf{B} $$ The right hand side still make sense when $mathbf{A}=mathbf{C}=nabla$. $$ nablatimes(nablatimesmathbf{B})=nabla(nablacdotmathbf{B})-(nablacdotnabla)mathbf{B}+color{red}text{correction} $$ We note that we didn't push any vector field pass a $nabla$, so there are no correction terms. (We did change the order of $mathbf{C}$ and $mathbf{A}$ in $mathbf{A}(mathbf{C}cdotmathbf{B})$ but they are both the differential operator $nabla$ so the symmetry of partial derivatives means there are no correction term).

However, I'd seriously advice against doing this eyeballing for anything more complicated. To see why, think about $(mathbf{A}timesnabla)timesmathbf{B}=mathbf{A}cdotnablamathbf{B}-mathbf{A}(nablacdotmathbf{B})=mathbf{A}times(nablatimesmathbf{B})+(mathbf{A}cdotnabla)mathbf{B}-mathbf{A}(nablacdotmathbf{B})$.

There are many identities in vector calculus that can be referred to simplify such expressions.

Using $nablacdot(phimathbf A)=phinablacdotmathbf A+(nablaphi)cdotmathbf A$, which looks analogous to the product rule of differentiation, you get $$nabla.(phivecnablapsi)=phinabla^2psi + (vecnablaphi).(vecnablapsi)$$

EDIT:

Consider ${color{red}{mathbf C}}times(mathbf Atimesmathbf B)=mathbf A(mathbf {color{red}{mathbf C}}cdotmathbf B)-mathbf B(mathbf {color{red}{mathbf C}}cdotmathbf A)$ and ${color{red}{mathbfnabla}} times (mathbf{A} times mathbf{B}) = mathbf{A}({color{red}{mathbfnabla}} {cdot} mathbf{B}) ,-, mathbf{B}({color{red}{mathbfnabla}} {cdot} mathbf{A}) ,+, (mathbf{B} {cdot} nabla) mathbf{A} ,-, (mathbf{A} {cdot} nabla) mathbf{B}$ .

Where is the analogy? I think after deriving a few formulae listed in the link attached, you could tell apart where the analogy works and where it doesn't.

You are computing the divergence of the vector field $left(phi frac{partial psi}{partial x_i}right)_{i=1,cdots,n}$, so you just get $$ sum_{i=1}^n frac{partial}{partial x_i} left(phi frac{partial psi}{partial x_i}right) $$

using the product rule you simply get $$ sum_{i=1}^n left(frac{partial phi}{partial x_i} frac{partial psi}{partial x_i} + phi frac{partial^2 psi}{partial x_i^2}right) = nabla phi cdot nabla psi + phi nabla^2psi $$

once you know the result, you can "build" some mnemonic related to the product rule, but you still need to know what first and second order operators you must use.