The Gimli non-linear operator
a bit late, if I had been pinged on this, I could have answered earlier.

$x \vee z$ was chosen for the following reason:

$a \wedge b$ is biased towards 0. As a result if we only used $\wedge$ the linear weight of the equation would have a strong bias towards 0.
As $a \vee b$ is biased towards 1, this aims to compensate.


The $z ≪ 1$ instead of just $z$ in the equation is necessary to ensure that this is a permutation, otherwise the equations do not have an inverse and we really wanted a permutation and not a transformation.

As for the why $⊕⊕((∧)≪3)$, if you remove the $x \leftrightarrow z$ swap, you realize that the shape of the expression is:
$x \leftarrow x \oplus z' \oplus EXPR$
$y \leftarrow y \oplus x \oplus EXPR$
$z \leftarrow z \oplus y \oplus EXPR$
We already discussed why $z'$ is $z ≪ 1$, now if you look at $EXPR$ you notice that in all cases, the destination is not included in the non-linear binary expression, as a result, this increase the complexity of linear trails as you have to track more variables.