In what follows $U(z_0,r) = \{z \in \mathbb{C} : |z-z_0| < r \}$ where $|.|$ is the Euclidean norm, $z_0 \in \mathbb{C}$ and $r > 0$.
$\textbf{Problem statement}:$
Let $B$ and $D$ be open subset of $\mathbb{C}$, and suppose both $B$ and $D$'s boundaries consist of finitely many piecewise analytic curves, where a piecewise analytic curve is a (continuous) mapping $\gamma : [0,1] \rightarrow \mathbb{C}$, which on each (closed) subinterval $[t_{j-1},t_j]$ of some partition $t_0 = 0 < t_1 < ... < t_n = 1$, is the restriction of a holomorphic function defined on a neighbourhood of the subinterval, moreover we may suppose that this holomorphic function has non-vanishing derivative at each point of the subinterval.
Let $\phi$ be a univalent (i.e. holomorphic and one to one) mapping of $B$ onto $D = \phi(B)$ and let $H$ be the set of points of $\partial B$ where $\partial B$ fails to be analytic or is mapped to by $\phi$ to a point of $\partial D$ where $\partial D$ fails to be analytic.
Suppose $E$ is a compact subset of $\overline{B} \setminus H$ such that every point $z_0 \in E$ has a neighbourhood $U(z_0,r)$ such that either $U(z_0,r) \cap B$ is connected or $U(z_0,r) \subset \overline{B}$.
Now let $z_0 \in E \cap \partial B$ and assume that for some neighbourhood $U(z_0,r)$ the intersection $U(z_0,r) \cap B$ is connected. Then $\phi$ extends to an analytic function in a whole open neighbourhood of $z_0$ (say, for concreteness, to a small open disk centred at $z_0$).
$\textbf{My attempt and specific question}$ : It seems clear that what is going on (or ought to be) is that, there is a analytic $\textbf{one-sided}$ subarc of the boundary of $B$, which also forms part of the boundary of $B \cap U(z_0,r)$ that $\phi$ maps to another analytic subarc of the boundary of $D$. Therefore we can apply the usual Schwarz reflection principle to produce the extension (after pre and postcomposing with analytic mappings of both subarcs, mapping to the real-line).
What is unclear to me is how we can show that we are in precisely this situation, that is how can I show that such an analytic subarc of the boundary of $B$ must be one-sided? It seems connectedness of $U(z_0,r) \cap B$ is crucial here, since if we had another analytic arc of the boundary of $B$ crossing $z_0$, then we would not be in such a situation.
By $\textbf{one-sided}$, I mean the following: There is a univalent function $\phi : U(0,1) \rightarrow \mathbb{C}$ with $\phi(0) = z_0$, such that $\phi(z)$ lies inside $B$ when $z$ lies in the upper half of the open unit disk, and outside $B$ (i.e. in the exterior of $B$) when $z$ lies in the lower half of the open unit disk, and $\phi(-1,1)$ is the analytic subarc of $B$ mentioned in the previous paragraph.