Is there any triangle-free, star-cutset-free, circle graph, with more than n edges?
Suppose $G$ is a triangle-free star-cutset-free circle graph.
I will show that $G$ contains no vertex with degree more than 2. Therefore, $G$ has at most $n$ edges.

Consider a circle representation $C$ of $G$. A set of chords is parallel if no two of them cross but there is a line crossing all the chords.

Property 1: $C$ has no 3 parallel chords.

Proof. Suppose $C$ has 3 parallel chords. Condider the vertex $v$ corresponding to the middle chord. Then, $N[v]$ is a cutset. This proves the property.

For the sake of contradiction, assume $G$ has a vertex $v$ of degree at least 3. Then, the chord corresponding to $v$ intersects 3 other chords. Since these 3 chords intersect one line, they are either parallel or two of them intersect. Due to Property 1, two of them intersect, which means their vertices form a triangle with $v$, which contradicts $G$ being triangle-free.