$$\text{When looking at the order }\mathbb{Z}\setminus\{0\},\unrhd with $$
$$x\unrhd y \Leftrightarrow \left\{ \begin{array}{@{}ll@{}} x\leq y \text{ if x$\cdot$y < 0} \\ |x|\leq |y| \text{ if x$\cdot$y > 0} \end{array}\right.$$
is it correct to interpret this as follows:
when $x \cdot y$ < 0, one (and only one) of both should be negative therefor $x\leq y$ (or $y\leq x$ in wich case switch x and y) in other words a negative number is always smaller than a positive number.
When $x \cdot y$ > 0 both numbers have the same sign (both positive or negative) interpreting this half of the order as: the one with the smallest absolute value is smallest so, the one closest to zero.
In my mind this order behaves as the regular "$\leq$ " with the sole difference that, for instance, $$-2 \leq -6 $$ because the absolute value of -6 is bigger than that of -2.
Knowing whether i interpreted this order right or wrong will be key in solving a linked problem.
$$ $$
Your interpretation is correct. The subset of negative integers is ordered as if it was specular with respect to the one of positive integers.