I am not sure that the following qualifies for a simple and intuitive explanation, but it may shed some light from a probabilistic perspective in a more general context of random trees:

as in the Dershowitz and Zaks' paper, the number returns of a random uniform Dyck path of semilength $𝑛$ to the $𝑥$-axis has the same distribution as the outdegree of a uniform random plane tree with $2n$ edges.

it is well known that a uniform random plane tree with $2n$ edges has the same distribution as a Bienaymé-Galton-Watson random tree with geometric $1/2$ offspring distribution, conditioned to have $2n$ edges.

Now consider a more general critical offspring distribution $\mu$ (a probability distribution on the non-negative integers with mean $1$), and assume that: $\mu(0)+\mu(1)<1$ (to avoid trivial cases), $\mathbb{P}(T \text{ has } n \text{ vertices})>0$ for every $n$ sufficiently large (to avoid periodicity) and that the variance of $\mu$ is finite (these assumptions are satisfied for the geometric $1/2$ offspring distribution).

Let $T$ be a non-conditioned Bienaymé-Galton-Watson tree with offspring distribution $\mu$ (roughly speaking, one starts with an ancestor which has a random number of children distributed according to $\mu$, then each one of its children has an independent random number of children distributed according to $\mu$, and so on) and let $T_n$ be a Bienaymé-Galton-Watson tree with offspring distribution $\mu$ conditioned on having $n$ vertices (which is well defined for $n$ suffficiently large).

By using the coding of BGW trees by the so-called Lukasiewicz path (see e.g. Section 1.1 here), if $(S_n)_{n \geq 0}$ is a random walk on the integers starting from $0$ and with jump distribution given by ${S}_1=\mu(i+1)$ for $i \geq -1$, and using the cyclic lemma (see e.g. Section 5 here), one gets that

\begin{eqnarray*}
\mathbb{P}(\text{degree root}(T_n) =k) & = \mu(k) \frac{\mathbb{P}(\text{a random forest with } k \text { trees has } n-1 \text{ vertices})}{\mathbb{P}(T \text{ has n vertices})} \\
& = \mu(k) \frac{ \frac{k}{n-1} \mathbb{P}(S_{n-1}=-k)}{\frac{1}{n}\mathbb{P}(S_n=-1)}.
\end{eqnarray*}

By the local limit theorem, there exists a constant $C>0$ such that $\frac{\mathbb{P}(S_{n-1}=-k)}{\mathbb{P}(S_n=-1)} \leq C$ for every $n$ sufficiently large and $k \geq 1$, so that for every $n$ sufficiently large and $k \geq 1$
$$ \mathbb{P}(\text{degree root}(T_n) =k) \leq C k \mu(k).$$
It follow that $\mathbb{E}[\text{degree root}(T_n)]$ is bounded since $\mu$ has finite variance.

Also, another reason that the expected number of returns of a Dyck path of semilength $𝑛$ to the $𝑥$ tends to $3$ is that $3$ is the expectation of a size-biased geometric $1/2$ random variable. Indeed (see this survey for details):

Large random Bienaymé-Galton-Watson trees with critical offspring distribution converge in distribution for the local topology to the so-called Kesten's infinite random tree.

In this infinite random tree, the outdegree of the root follows a size biased distribution.

As a consequence, the number of returns of a Dyck path of semilength $𝑛$ to the $𝑥$-axis converges in distribution to a size-biased geometric $1/2$ random variable, and one can see that this convergence also holds in expectation.