Let $\mathrm{CC}_n$ denotes the set of Catalan codes of length $n$, i.e.
$$\mathrm{CC}_n = \left\{ (c_0, \dots, c_{n-1})\in \mathbb{Z}_{\geq 0}^n\ :\ c_0+\dots+c_i\leq i\ \text{for all}\ i=0,1,\dots,n-1 \right\}.$$
For $c\in \mathrm{CC}_n$, denote $s(c) = n-c_0-\dots-c_{n-1}$.

Then
$$N(h_n) = \sum_{k=0}^n [x^k]\ \sum_{c\in \mathrm{CC}_{n-k}} (1-2x)^{-s(c)-1} \cdot\prod_{i=0}^{n-k-1} \frac{1+(1-2x)^{-c_i-1}}{2},$$
where $[x^k]$ is the operator taking the coefficient of $x^k$. (I did not think much about simplifying this expression.)

Notice that there is a similar expression for $N(g_n)$, which indeed simplifies to the large Schröder number $S_n$:
$$
\begin{split}
N(g_n) &= \sum_{k=0}^n [x^k]\ \sum_{c\in \mathrm{CC}_{n-k}} (1-x)^{-s(c)-1} \cdot\prod_{i=0}^{n-k-1} (1-x)^{-c_i-1} \\
& = \sum_{k=0}^n [x^k]\ C_{n-k}\cdot (1-x)^{-2(n-k)-1} \\
& = \sum_{k=0}^n C_{n-k}\cdot \binom{2n-k}{k} \\
& = S_n.
\end{split}$$

This is my SAGE code implementing the aforementioned formula for $N(h_n)$:

```
def myNh(n):
R.<x> = PowerSeriesRing(ZZ)
return sum( \
sum( \
( prod( \
1 + (1-2*x+O(x^(k+1)))^(-c[i]-1) \
for i in range(n-k) ) \
) * (1-2*x+O(x^(k+1)))^(-(n-k-sum(c))-1) / 2^(n-k) \
for c in map(lambda x: x.to_Catalan_code(), DyckWords(n-k)) )[k] \
for k in range(n+1) )
```

For $n=0,1,\dots,14$, it gives

1, 3, 11, 46, 210, 1018, 5150, 26889, 143829, 784167, 4341843, 24348352, 138007784, 789375504, 4550522248

which match the values computed by others in the comments. I've added it as the sequence A281548 to the OEIS.