In THIS ANSWER, I discussed the meaning of the identity
$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag 1$$
In that expression, the equality is interpreted as a set equality. This means for any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$. In addition, the sum of any values of $\log(z_1)$ and $\log(z_2)$ can be expressed as some value of $\log(z_1z_2)$.
Now, suppose $f(z)=\sqrt{z^2-1}=\sqrt{(z-1)(z+1)}$. By definition, we have
$$\begin{align}
f(z)&=e^{\frac12\log((z-1)(z+1))}\\\\
&=e^{\frac12\left(\log(z-1)+\log(z+1)\right)}\\\\
&=e^{\frac12\sqrt{z-1}}e^{\frac12\sqrt{z+1}}\\\\
\sqrt{z^2-1}&=\sqrt{z-1}\sqrt{z+1}\tag2
\end{align}$$
The equality in $(2)$ is a set equivalence analogous with $(1)$.
EXAMPLE:
For the example given in the OP, $z=-2$. We denote by $z_1$ and $z_2$, $z_1=z+1$ and $z_2=z-1$. Clearly, $z_1=-1$, $z_2=-3$, and $z_1z_2=3$.
The multi-valued term $\log(z_1z_2)$ is given by
$$\log(z_1z_2)=\log(3)=\log(|3|)+i2n\pi\tag 3$$
for any integer $n$. If we define $\log(z_1)=i\pi$ and $\log(z_2)=\log(|3|)+i\pi$, and if $n=1$ in $(3)$, then $\log(z+1)+\log(z-1)=\log(z^2-1)$. However, for any other $n$, the equality does not hold.
If we use $(3)$ to calculate $\sqrt{z^2-1}=\sqrt{z_1z_2}$, then we obtain
$$\sqrt{z_1z_2}=\sqrt{3}=\sqrt{|3|}e^{in\pi}\tag 4$$
For $n$ odd, the equality $\sqrt{z_1z_3}=-\sqrt{|3|}=\sqrt{z_1}\sqrt{z_2}$ holds, while for $n$ even, the equality does not hold.
Now, let's be more general. We cut the plane using the principal branch for $\sqrt{z-1}$ and observe that
$$\sqrt{z-1}=\sqrt{|z-1|}e^{i\text{Arg}(z-1)/2}$$
where $-\pi<\text{Arg}(z)\le \pi$ is the principal argument of $z$.
Similarly, using the principal branch for $\sqrt{z+1}$, we see that
$$\sqrt{z+1}=\sqrt{|z+1|}e^{i\text{Arg}(z+1)/2}$$
for $-\pi<\text{Arg}(z)\le \pi$.
If we wish to use the principal branch for $\sqrt{z^2-1}$, then the equality $\sqrt{z^2-1}=\sqrt{z-1}\sqrt{z+1}$ does not hold in general. Rather, the equality holds as follows:
If $-\pi<\text{Arg}(z-1)+\text{Arg}(z+1)\le \pi$, then $\sqrt{z^2-1}=\sqrt{|z^2-1|}e^{i\text{Arg}(z^2-1)}$.
If $\pi<\text{Arg}(z-1)+\text{Arg}(z+1)\le 2\pi$, then $\sqrt{z^2-1}=\sqrt{|z^2-1|}e^{i\text{Arg}(z^2-1)+2\pi}$
If $-2\pi<\text{Arg}(z-1)+\text{Arg}(z+1)\le -\pi$, then $\sqrt{z^2-1}=\sqrt{|z^2-1|}e^{i\text{Arg}(z^2-1)-2\pi}$
More simply expressed, the equality holds by using the expression $\arg(z^2-1)=\text{Arg}(z-1)+\text{Arg}(z+1)$.
Let's use the previous example in which $z=-2$. We find that $\text{Arg}(z-1)=\text{Arg}(z+1)=\pi$ and $\text{Arg}(z^2-1)=0$. Indeed, we have
$$\sqrt{z-1}\sqrt{z+1}=\sqrt{z^2-1}$$
when we take $\arg(z^2-1)=\text{Arg}(z^2-1)+2\pi=0+2\pi=2\pi$.
More simply, we have equality with $\arg(z^2-1)=\text{Arg}(z-1)+\text{Arg}(z+1)=2\pi$.
