deep breaths and calming thoughts
Let's not worry about the symbols $\sqrt{}$ and $\pm$ but lets address the core issue.
If you have a ration $\frac xy$ (assume $x > 0;y>0$) what are its square roots; and are the set of square roots of $\frac xy$ and the ratios of the square roots of $x$ and $y$ the same thing.
The answer is yes. If $x$ has two square roots $a$ and $-a$ (assume $a > 0$) and $y$ has two square roots $b$ and $-b$ (assume $b > 0$) then:
The two square roots of $\frac xy$ are $\frac ab$ and $-\frac ab$ as $(\frac ab)^2 = (-\frac ab)^2 = \frac {a^2}{b^2} = \frac xy$.
Furthermore $\frac {-a}{b} = -\frac ab$; $\frac {a}{-b} = -\frac {a}b; \frac {-a}{-b} = \frac ab$ and the fourth ratio is $\frac ab$.
So the square roots of $\frac xy$ and the ratios of the square roots of $x$ and the square roots of $y$ are the same two numbers.
...
So now let's bring in the $\sqrt{}$ symbol.
If $w > 0$ then $\sqrt{w}$ is the non-negative (positive because $w > 0$) number $v$ so that $v^2 = w$, that is to say the positive square root. There is also another negative square root that happens to be $-v$ and we right that as $-\sqrt w$.
Note: $\sqrt{w}$ is always and always will be the NON-negative square root, whereas there will (if $w > 0$) be two square roots; one negative the other positive.
So $\sqrt{xy} =\frac {\sqrt x}{\sqrt y}$. And using your example $\sqrt{\frac {16}{4}} = \frac {\sqrt{16}}{\sqrt{4}} = \frac 42$. That's all there is too it becase square roots are not negative.
But what of the other square roots?
$\frac {16}{4}$ has two square roots: $\sqrt{16}{4} = \frac 42$ and $-\sqrt{16}{4} = -\frac 42$.
$16$ has two square roots: $\sqrt 16 =4$ and $-\sqrt {16}=-4$ and $4$ has two square roots: $\sqrt{4} =2$ and $-\sqrt 4 = -2$. So there are four possible combination of ratios.
$\frac {\sqrt{16}}{\sqrt {4}} = \frac 42$
$\frac {-\sqrt{16}}{\sqrt {4}} = -\frac 42$
$\frac {\sqrt{16}}{-\sqrt {4}} = -\frac 42$
and $\frac {-\sqrt{16}}{-\sqrt {4}} = \frac 42$
Those are the two values.
And that's that.
....
Okay what about the symbol $\pm$.
So we can condense the above by:
$\frac {\pm 4}{\pm 2}$ and that can be any of the four $\frac {4}{2},\frac {-4}{2},\frac {4}{-2},\frac {-4}{-2}$.
But as $\frac {4}{-2} = \frac {-4}{2}$ and $\frac {-4}{-2} = \frac {4}{2}$ we can "fix" the numerator to be positive and simply write $\frac {\pm 4}{2}$ (although it'd probably by better to write it as $\pm \frac 42$.
....
So what about this "breaking" rule?
Well, ther's really no such thing. If the numerator can be one or the other and the denominator can be one or the other there is now reason you can break them.
What matters is if the terms are dependant upon each other. So as $\pm(a-b)$ This is eithere $a-b$ or $-a + b$. So the first term is $\pm a$ and the second term is $\mp b$. But they depend on each other. So we can't "break" them".
But that would not be the case if , say, we were asked $x^2 = 9$ and $y^2 = 4$ what is $x + y$. Well, $x = \pm 3$ and $y = \pm 2$ so $x+y$ maybe any of the four values $\pm x \pm y$ ($3+2; 3-2; -3+2; -3-2$). And we certainly can break them as they are completely independent of each other.