Here's a solution.

**1. Number of touches of diagonal (more simple case).**

Consider generating function of lattice paths which don't touch and don't cross the diagonal:
$$
Cat(z) = \dfrac{1 - \sqrt{1 - 4z}}{2} = z + z^2 + 2z^3 + 5z^4 + 14z^5 + \ldots
$$
Then we take product of $ k $ Catalan generating series which stands for lattice paths touching or crossing exactly $ k$ points:
$$
[z^n] (Cat(z))^k = \text{# lattice paths from $(0,0)$ to $(n,n)$ having exactly k diagonal points}
$$
Then we assemble everything into bivariate generating function, adding multiple $2^k$ for two different possible sides of Catalan paths:
$$
F(z,u) = \sum_{k \geq 1} 2^k(Cat(z))^k u^k = \dfrac{1}{1 - u(1 - \sqrt{1-4z})}
$$
Note that if we plug $u=1$ then we obtain all possible paths from $(0,0)$ to $(n,n)$ regardless of number of diagonal crosses:
$$
\dfrac{1}{\sqrt{1-4z}} = 1 + 2z + 6z^2 + 20z^3 + \ldots = \sum_{n \geq 0} {2n \choose n} z^n \enspace .
$$
According to the method of bivariate generating functions, the expectation is expressed by applying the point derivative operator:
$$
u \dfrac{d}{du} \sum_{n, k, \geq 0} a_{n,k} z^n u^k =
\sum_{n, k, \geq 0} k a_{n,k} z^n u^k \enspace ,
$$
Denoting by $[z^n]f(z)$ the coefficient at $z^n$ in function $f(z)$, and by $U_n$ – random variable denoting the number of crossings, we obtain
$$
\mathbb E U_n = \dfrac{[z^n] \left.u \dfrac{d}{du} F(z,u)\right|_{u=1}}{[z^n] \left.F(z,u)\right|_{u=1}} \enspace .
$$
$$
\left.\dfrac{d}{du}\dfrac{1}{1 - u(1 - \sqrt{1-4z})} \right|_{u=1} =
\left.\dfrac{1 - \sqrt{1-4z}}{\left[1 - u(1 - \sqrt{1-4z})\right]^2} \right|_{u=1} =
\dfrac{1 - \sqrt{1-4z}}{1 - 4z}
$$
Applying asymptotic extraction of coefficients,
$$
\mathbb E U_n =
\dfrac{[z^n] \dfrac{1}{1 - 4z}}
{[z^n] \dfrac{1}{\sqrt{1 - 4z}}} =
\dfrac{4^n}{2n \choose n } \sim \sqrt{ \pi n}
$$

**2. Number of ***crossings* of diagonal (the same idea, the functions more clumsy).
Instead of non-touching Catalan we consider non-crossing non-empty Catalan:
$$
Cat^\ast(z) = \dfrac{1 - \sqrt{1 - 4z}}{2z} - 1 = z + 2z^2 + 5z^3 + 14z^4 + \ldots
$$
Then we multiply Catalan generating functions in an alternating manner (times 2 for initial one):
$$
[z^n] 2 (Cat^\ast(z))^k
$$
and doing the same stuff as before, we arrive to bivariate gf (minus one because the board $0 \times 0$ was counted twice):
$$
G(z,u) = 2 \cdot \dfrac{1}{1 - u \cdot \dfrac{1 - 2z-\sqrt{1 - 4z}}{2z}} - 1
$$
Check what happens at $u=1$:
$$
G(z,1) = \dfrac{4z}{4z - 1 + \sqrt{1 - 4z}} = 1 + 2z + 6z^2 + 20z^3 + \ldots
$$
After differentiation and plugging $u=1$ obtain the asymptotics:
$$
\left. u \dfrac{d}{du} G(z,u) \right|_{u=1} =
\dfrac{-2z - \sqrt{1-4z}+1}{z \left(1 - \frac{-2z-\sqrt{1-4z}+1}{2z}\right)^2}
$$
Near the singularity $z = \frac14$ we have an expansion:
$$
\left. u \dfrac{d}{du} G(z,u) \right|_{u=1} \sim \dfrac{1}{2 (1 - 4z)}
$$
Finally, the asymptotics is as in the first case, divided by 2:
$$
\mathbb E U_n \sim \dfrac{\sqrt{\pi n}}{2}
$$

P.S. The final answer looks familiar to the expression of T.Amdeberhan which is of order $ \dfrac{1}{2}\cdot \dfrac{4^n}{ {2n \choose n}}$