As for the second question: "Why are we allowed to make this manipulation in the special case of $\sqrt{-a}=\sqrt{(-1)\times a}=\sqrt{-1}\times \sqrt{a}=i\sqrt{a}$ when it otherwise isn't allowed"
tldr: it is a result of the rigorous definition of square roots in the context of complex analysix
This is unfortunately a matter of convention as to what we mean by $i$ in the first place. We are often told things like $i$ is the complex number such that $i=\sqrt{-1}$. Other people might say things like $i$ is a complex number with the property that $i^2=-1$. The more rigorous people often use this second one, but notice that we can have two different imaginary units like this, $i$ and $j$ where $j=-i$. The sad truth is that there really isn't much of any difference between these, but we just agreed that one of them is what we will call $i$ and that will be in the "positive imaginary direction" while the other is in the "negative imaginary direction."
Now... we need to talk about what we actually mean by $\sqrt{~~}$. We have that $\sqrt{x}$ for positive real number $x$ is the positive real number such that when you square it you get $x$. That is to say $(\sqrt{x})^2=x$, but this doesn't necessarily make sense when we put something other than a real number into it.
To define it for complex numbers (and negative numbers), we turn to the polar representation of complex numbers. $z = r\cdot e^{i\theta}$ where $r$ is the magnitude of the number and $\theta$ is the direction on the complex plane that the number is in. We will say that a complex number is in standard form if $r$ is non-negative real and $\theta$ is a real number in the range $0\leq \theta<2\pi$.
We can then define $\sqrt{z}$ as $z^{\frac{1}{2}}$ which becomes $(re^{i\theta})^{\frac{1}{2}}=r^{\frac{1}{2}}e^{i\theta/2}$, again emphasizing that we are forcing the values of $\theta$ to be such that $0\leq \theta<2\pi$. Note then that $0\leq \theta/2<2\pi$ will make it so this expression for $\sqrt{z}$ is also in standard form. By insisting things be in standard form, we successfully make it so that this is indeed a function and not a multivalued function, having it so that there is a single output. In fact, the image of the complex square root function results only in values of $\theta$ such that $0\leq \theta <\pi$. This is part of why we say $\sqrt{4}=2$ and only $2$, not $-2$, because $-2$ has $\theta$ values of $-\pi$ which is outside of the range of the image.
What the $\sqrt{~~}$ as defined above accomplishes then, is it takes an input as a complex number, and rotates it half of the way towards the positive real axis from its current location. It then also changes the magnitude of the number into the real square root of the magnitude of the number.
In the special case of negative real numbers, the effect of the rotation has the result pointing in the positive imaginary direction. Hence $\sqrt{-a}=i\sqrt{a}$ for positive real $a$.
In fact, by looking at the square root in this more rigorous fashion, we can explain why many of these unusual counterexamples happen. $\sqrt{-1}\times \sqrt{-1}$ the $\theta$ values for each are originally $\pi$, the square rooting turns them each into $\pi/2$, and the multiplication of the numbers adds them, putting it back to $\pi$ again. However $\sqrt{(-1)\times (-1)}$ the theta value in standard form is $0$ instead of $2\pi$, so halving the theta value makes it $0$, not $\pi$.
