Consider a circle with a circumference of $n$. On this circle, I define two arcs of length $k<n$, $A_1$ and $A_2$. The centres of the two arcs are uniformly distributed on the circle.
Let $\Omega_{1}=A_1 \setminus A_2$ and $\Omega_{2}=A_2 \setminus A_1$ such that the length of $\Omega_1 \cup \Omega_2$ is $2k$ minus the overlapping part of the two arcs.
What is the expectation of the length of $\Omega_1 \cup \Omega_2$?
By rotational symmetry, you can view the location of the first arc as fixed, and consider only the randomness of the second arc's position relative to the first one.
If the [absolute] arc distance between the centers of the two arcs is $X$, then you can check that the length of $\Omega_1 \cup \Omega_2$ is $2 \min\{X,k\}$. Then, note that $X$ is uniformly distributed between $0$ and $n/2$, so $$E[2 \min \{X, k\}] = 2 \int_0^{n/2} \frac{1}{n/2} \min\{t, k\} \, dt.$$ I'll leave you to compute this integral, which will have different forms depending on whether $k \le n/2$ or $k \ge n/2$.