Now, depending on the values of $y$, we have $\lvert y-1 \rvert = \pm(y-1)$; and similarly, $\lvert x+3 \rvert = \pm(x+3)$. Thus, $(6)$ can be written as $$y-1 = \pm C_1(x+3) \qquad \text{or} \qquad y= 1\pm C_1(x+3)$$
It feels like it should be obvious to me but I'm not quite seeing how the two expressions are equal. I suppose that the line of thinking is exactly like $|x| = \pm x$?
It's a request for its formal definition, no?
You can define absolute value piecewise:
$$ |x| = \begin{cases} x & x \geq 0 \\ -x & x < 0\end{cases} $$
or you can define it as $ |x| = \sqrt{x^2} $
In either case, it can be reinterpreted as two cases: a positive or a negative value of the argument $ x $, depending on the sign of $ x $.
Reading a little closer makes me think you are asking the identical question of why $ | x - 3 | = \pm (x - 3) $. You just call on algebra to say that: let $ z = x - 3 $, and then it becomes just the two cases of the sign of $ z $.
Another thought that's a pretty primitive one is to interpret simple absolute value equations with a number line. You can think of $$ |x - p| \leq d $$ (which is a bit different, it is an inequality) as the points $ p $ on a number line that are within distance $ d $ of $ x $. Putting in an equals sign just means the two edges points.
This post on this site shows the picture I mean in one of its answers: link
All of it is a bit of semantics, just saying it is two cases for positive and negative, $ \pm $, means the same thing any way you slice it.