Write $f$ in the form
$$f(z):={z\over i}\>{\rm pv}\sqrt{1-{1\over z^2}}\qquad\bigl(z\in\Omega:={\mathbb C}\setminus[{-1},1]\bigr)\ .$$
It is easily checked that for $z\in\Omega$ the expression $1-{1\over z^2}$ cannot be real $\leq0$. It follows that for $z\in\Omega$ the principal value ${\rm pv}\sqrt{1-{1\over z^2}}$ and therewith $f$ is well defined and analytic in $\Omega$.
The principal value $w\mapsto{\rm pv}\sqrt{w}$ has the symmetry property $${\rm pv}\sqrt{\bar w}=\overline{{\rm pv}\sqrt{w}}$$ on its domain of definition. It follows that $f$ has the symmetry properties
$$f(\bar z)={\bar z\over i}{\rm pv}\sqrt{1-{1\over \bar z^2}}=-\overline{f(z)}\ ,\qquad f(-z)=-f(z)\qquad(z\in\Omega)\ .\tag{1}$$
Writing $z=x+iy$ one has
$$f(x-iy)=-\overline{f(x+iy)}\ ,$$
and this shows that the second of your conjectures only is valid at the points $z\in\Omega$ where $f$ happens to asssume a real value. (These are the points $\ne0$ on the imaginary axis.)
In order to discuss the $\lim f(x+iy)$ for $y\to0+$ and $y\to0-$ when $|x|<1$ we shall argue as follows: The value $f(i)=\sqrt{2}$ coincides with the principal value of the originally given expression $\sqrt{1-z^2}$ at $i$. Furthermore $1-z^2=1-x^2+y^2-2ixy$ is not real $\leq0$ in the open upper halfplane $H_+$: It can only be real on $H_+$ when $x=0$, and in this case $1-z^2=1+y^2>0$. It follows that in fact
$$f(z)={\rm pv}\sqrt{1-z^2}={\rm pv}\sqrt{1-x^2+y^2-2ixy}\qquad(z\in\Omega\cap H_+)\ .$$
This immediately implies
$$\lim_{y\to0+} f(x+iy)=\sqrt{1-x^2}\qquad\bigl(|x|<1)\ .$$
From the second identity $(1)$ we then infer that
$$f(z)=-{\rm pv}\sqrt{1-z^2}\qquad(z\in\Omega\cap H_-)\ ,$$
and in particular
$$\lim_{y\to0-} f(x+iy)=-\sqrt{1-x^2}\qquad\bigl(|x|<1)\ .$$
Therefore the first of your conjectures is correct.
