Let $n ≥ 3$. Find the number of NE lattice paths from $(0, 0)$ to $(n, n)$ that touch the diagonal $y = x$ at least twice

Well done. Notice that the second symmetry is not $2C_{n-1}$ but $4C_{n-1}$ because when you touch the line $x=y$ you either rebound or go to the other side. That gives $2$ possibilities and then you can either start below or above, that gives $2$ so $4.$ In that way you have $$\binom{2n}{n}-2C_{n-1}-4C_{n-1}=\binom{2n}{n}-6C_{n-1}.$$

There are $2C_{n-1}$ paths that never touch the diagonal between the endpoints. A path that touches the diagonal exactly once between the endpoints, at $\langle k,k\rangle$, is the union of a Dyck path of length $k-1$ and a Dyck path of length $n-k-1$, and there are $2$ choices for each of these paths, one above and one below the diagonal. Thus, there are

$$4\sum_{k=1}^{n-1}C_{k-1}C_{n-k-1}=4\sum_{k=0}^{n-2}C_kC_{n-2-k}=4C_{n-1}$$

paths that hit the diagonal exactly once between the endpoints, and the desired number is therefore

$$\binom{2n}n-6C_{n-1}\,.$$

As a quick minimal sanity check, for for $n=2$ this is $\binom42-6C_1=0$, and for $n=3$ it is $\binom63-6C_2=20-6\cdot 2=8$, both of which are correct.