Is it possible to extract a neighborhood around any point on a simple closed curve such that the boundary of this neighborhood intersects the curve at only two points?
For a simple closed curve $\gamma$, for any $z_0$ on $\Gamma$, there exists a neighborhood $U$ containing $z_0$ such that $\partial U$ has only two intersection points with $\Gamma$.
In the context of complex analysis, this assumption is often referred to as a 'free boundary arc'.
Considering the complex plane only.
The answer is positive. For any Jordan curve, there is a homeomorphism of the Riemann sphere, which sends this curve to a circle. Your statement immediately follows.
One proof of this statement involves the Riemann mapping theorem and the boundary correspondence theorem of Caratheodory. But there are also purely topological proofs, of course.
As a more direct "construction" of such a neighborhood one could proceed as follows.
Take any subarc $A$ of $\Gamma$ such that $z_0\in A$, but the endpoints $p,q$ of $A$ are different from $z_0$ (so $z_0$ is "between" $p$ and $q$ in $A$). Let $B=A\setminus\{p,q\}$ and $C=\Gamma\setminus B$. For each $b\in B$ let $U_b$ be the ball centered at $b$ of radius $\frac{d(b,C)}3$. Let $U=\cup_{b\in B}\,U_b$.
The verification that this works is easy, but here are some details. If $s$ is some point on the boundary of $U$, then there are sequences $s_n$ and $b_n$ such that $s_n\to s$ and $s_n\in U_{b_n}$.
Case 1. Some subsequence of the $b_n$ converges to an endpoint $p$ or $q$. Say without loss of generality $b_n\to p$. Then the radii of the balls $U_{b_n}$ converge to $0$ (since $p\in C$), hence also $s_n\to p$ and $s=p.$
Case 2. Some subsequence of the $b_n$ converges to some $b\in B$ (with $b\not\in\{p,q\}$). Without loss of generality $b_n\to b$. Then $s$ is on the circle which is the boundary of $U_b$ and the latter does not intersect $C$.
(So, the two intersection points could be prescribed in advance.)
