Solving $|a| < |b|$
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 $1$.
There are different approaches; one is to look at the zeroes of the expressions inside the absolute values, and split up $\mathbb R$ into intervals accordingly. In your example $|x + 2| < |x - 4|$, the points of interest are at $x=-2$ and $x=4$. You can therefore consider three cases:
1. If $x \in (-\infty,-2)$, then $x+2 < 0$ and $x-4 < 0$, so the absolute values will reverse the signs of both. This gives: $$\begin{align} -(x+2) &< -(x-4) \\ x+2 &> x-4 \\ 2 &> -4 \end{align}$$ This is true for all $x$ in the interval.
2. If $x \in [-2,4)$, then $x+2 \geq 0$ so its sign is unaffected by the absolute value, but $x-4 <0$ so its sign will be reversed: $$\begin{align} x+2 &< -(x-4) \\ 2x+2 &< 4 \\ x &< 1 \end{align}$$ Combining this last inequality with the assumption that $x \in [-2,4)$, we see that any $x$ in $[-2,1)$ is valid.
3. Finally, if $x \in [4,\infty)$, neither expression's sign is reversed: $$\begin{align} x+2 &< x-4 \\ 2 &< -4 \end{align}$$ This is false for all $x$ in the interval.
Putting all the information together from the above three cases, we have $x \in (-\infty, 1)$.
Note: as some other contributors have mentioned, there are simpler ways to deal with your problem, such as viewing it geometrically. The method that I have shown above is more useful when the expressions are more complicated or when you have several absolute values; for example, an inequality like $3 |x^2-1|+|x-2|+|x^2-3x| > 5$.
We can see $|x-a|$ as a distance point $x$ from point $a$.
Now, the question with the above "definition" would be like:
For which $x$ values distance point $x$ from $-2$ be less than distance point $x$ from $4$?
Clearly, by drawing it maybe, you can observe that the answer is for all $x<1$.