So, we have to prove the average number of sections is $n+1$. Let each of the $\binom{2n}{n}$ combinations has equal probability to happen. The number of sections equals the number of occurrences of "$01$" and "$10$" plus one. Let $\xi_i, i=1,2,\dots 2n-1$ be random variables, $\xi_i$ that takes value $1$ iff we have $01$ or $10$ at $i$-th and $i+1$-th positions of the sequence, and $0$ - otherwise. We have $$ E \xi_i =\displaystyle\frac{2\binom{2n-2}{n-1}}{\binom{2n}{n}}=\frac{n}{2n-1}$$Now, the number of sections equals $\left(1+\sum_{i=1}^{2n-1}\xi_i \right)$. Thus, by linearity of expectation, $\displaystyle E\left(1+\sum_{i=1}^{2n-1}\xi_i \right)=n+1$.