I believe that your suggestion is correct:
$\newcommand{\C}{\mathbb{C}}$

**THEOREM.** Suppose that $U,V\subset\mathbb{C}$ are domains, and that $f:U\to V$ is a conformal isomorphism. Then $f'$ has a (single-valued) logarithm if and only if $f$ "maps the outer boundary of $U$ to the outer boundary of $V$". (I.e., if $\gamma\subset U$ is a simply closed curve parameterized in positive orientation, then $f\circ\gamma$ also has positive orientation.)

A special case is, of course, that the logarithm always exists when $U$ is simply-connected.

To prove the Theorem, note that the existence of the logarithm of $f'$ means precisely that, on any essential curve in $U$, the derivative does not wind around zero.

*Claim.* If $\gamma\subset\C$ is a simple closed curve, and $f$ is holomorphic and injective in a neighbourhood of $\gamma$, then the winding number of $f'\circ\gamma$ around zero is either $0$ or $-2$, according to whether $f\circ\gamma$ has the same or the opposite orientation to $\gamma$.

This proves the theorem.

The claim should be intuitively plausible. To prove it, consider the case where $\gamma$ and $f(\gamma)$ are both the unit circle. Then clearly, for $z\in\gamma$,
$$ \arg f'(z) = \arg f(z) - \arg(z) $$
if $f$ is orientation-preserving, while
$$ \arg f'(z) = -\arg f(z) - \arg(z)$$
if $f$ is orientation-reversing.

To reduce the general case to this observation, e.g. assume w.l.o.g. that the curve $\gamma$ is analytic, and change coordinates using the Riemann mapping theorem.

(I think there should be a direct analytic proof that does not require this change of variable, but I find the above argument rather intuitive.)