We have $|a| \lt |b|\,$ if any of these is true:
(i) $\,a$ and $b$ are $\gt 0$ and $a \lt b$
(ii) $a\lt 0$ and $b\ge 0$ and $-a \lt b$
(iii) $b \lt 0$ and $a \gt 0$ and $a \lt -b\,$
(iv) $\,a\lt 0$ and $b\lt 0$ and $-a\lt -b$. We can rewrite this as $b \lt a$.
Four cases! Not surprising, since eliminating a single absolute value sign often involves breaking up the problem into $2$ cases.
Sometimes, one can exploit the simpler $|a| \lt| b|\,$ iff $\,a^2\lt b^2$. But squaring expressions generally makes them substantially messier.
Added: With your new sample problem, squaring happens to work nicely. We have $|x+2| \lt |x-4|$ iff $(x+2)^2 \lt (x-4)^2$. Expand. We are looking at the inequality
$$x^2+4x+4 \lt x^2-8x+16.$$
The $x^2$ cancel, and after minor algebra we get the equivalent inequality $12x \lt 12$, or equivalently $x\lt 1$. The squaring strategy works well for any inequality of the form $|ax+b| \lt |cx+d|$.
But the best approach for this particular problem is geometric. Draw a number line, with $-2$ and $4$ on it. Our inequality says that we are closer to $-2$ than we are to $4$. The number $1$ is halfway between $-2$ and $4$, so we must be to the left of $1$.