I think the problem may be in the bijection. Consider instead the following bijection between (planted plane) trees with $n$ vertices and Dyck paths from $(0,0)$ to $(n-1,n-1)$: perform a depth-first search on the tree, moving $\uparrow$ in the Dyck path when you follow an edge away from the root, and $\rightarrow$ in the Dyck path when you follow an edge towards the root. Then a tree of height $h$ corresponds to a Dyck path which reaches $y=x+h-1$ but not $y=x+h$, so $A_{n,h}$ counts the Dyck paths from $(0,0)$ to $(n-1,n-1)$ which avoid the boundaries $y=x+h$ and $y=x-1$. By the referenced formula of Mohanty,
$$A_{n,h} = \sum_{k \in \mathbb{Z}} \left[\binom{2n-2}{n-1-k(h+1)} - \binom{2n-2}{n+k(h+1)}\right]$$
Then
$$\begin{eqnarray*}
B_{n+1,h-1} &=& A_{n+1,n+1} - A_{n+1,h-1} \\
&=& \frac{1}{n+1} \binom{2n}{n} + \sum_{k \in \mathbb{Z}} \left[ \binom{2n}{n+1+kh} - \binom{2n}{n-kh} \right] \\
&=& \frac{1}{n+1} \binom{2n}{n} + \sum_{k \geqslant 0} \left[ \binom{2n}{n+1+kh} - \binom{2n}{n-kh} \right] + \\&& \sum_{k > 0} \left[ \binom{2n}{n+1-kh} - \binom{2n}{n+kh} \right] \\
&=& \frac{1}{n+1} \binom{2n}{n} + \binom{2n}{n+1} - \binom{2n}{n} + \\&& \sum_{k \geqslant 1} \left[ \binom{2n}{n+1+kh} - \binom{2n}{n-kh} - \binom{2n}{n+kh} + \binom{2n}{n+1-kh} \right] \\
&=& \sum_{k \geqslant 1} \left[ \binom{2n}{n+1+kh} - \binom{2n}{n-kh} - \binom{2n}{n+kh} + \binom{2n}{n+1-kh} \right] \\
&=& \sum_{k \geqslant 1} \left[ \binom{2n}{2n-(n+1+kh)} - \binom{2n}{n-kh} - \binom{2n}{2n-(n+kh)} + \binom{2n}{n+1-kh} \right] \\
&=& \sum_{k \geqslant 1} \left[ \binom{2n}{n-1-kh} - 2\binom{2n}{n-kh} + \binom{2n}{n+1-kh} \right] \\
\end{eqnarray*}$$
as desired.
Here is a solution to your warm up problem. It uses a few known elementary identities, and some short inductions for identities I didn't recognize. I also changed your notation slightly from $l$ to $k$ in the internal summations.
Setting $x=2$, the first term vanishes, and we combine the second and third terms as $(B)$ and take the last term as $(A)$.
\begin{align}
&\sum_{j=0}^i \binom{n-1-j}{i-j} 2^{j-1} \sum_{k=0}^j (-1)^k \binom{n+1}{k} 2^{i-k-1}(i-k) \tag{A} \\
&\sum_{j=0}^{i-2} \binom{n-1-j}{i-2-j} 2^{j-1} \sum_{k=0}^j (-1)^k \binom{n+1}{k} \left( 2^{i-k} + 2^{i-k-1}(i-2-k)\right) \tag{B}
\end{align}
Fortunately for us, both are equal to $i\cdot2^{i-2}$ when $i$ is even, and do not depend on $n$.
First, we focus on $(A)$. Switch the order of summation to get
\begin{equation}
\sum_{k=0}^i (-1)^k \binom{n+1}{k} 2^{i-2}(i-k) \sum_{j=k}^i \binom{n-1-j}{i-j} 2^{j-k} \tag{A}
\end{equation}
Then reindex the internal summation, and repeatedly apply the hockey stick identity to show
\begin{equation}
\sum_{j=0}^{i-k}\binom{n-1-k-j}{i-k-j}2^j = \sum_{j=0}^{i-k}\binom{n-k}{j}
\end{equation}
I found it easier to write out by changing variables $a=i-k-j$, $b=i-k$ and $c=n-i-1$, then simplifying $\sum_{a=0}^b \binom{c+a}{c}2^{b-a}$.
$(A)$ naturally splits at the point $i-k$, and since we want to show that the whole thing is $i\cdot2^{i-2}$, we can break it into two pieces and show
\begin{align}
\sum_{k=1}^i (-1)^k k\binom{n+1}{k}\sum_{j=0}^{i-k} \binom{n-k}{j} &=0 \tag{A1}\\
\sum_{k=0}^i (-1)^k \binom{n+1}{k} \sum_{j=0}^{i-k} \binom{n-k}{j} &=1 \tag{A2}
\end{align}
We can simplify $(A1)$ slightly by incorporating $k$ into the binomial, and ignoring the $n+1$ term that comes out.
Switching the order of summations again, we write
\begin{equation}
Q(n,i) = \sum_{j=0}^{i-1} \sum_{k=1}^{i-j} (-1)^k \binom{n}{k-1} \binom{n-k}{j}
\end{equation}
We will induct on $n$ and $i$ to show that $Q(n,i)=0$ for $i$ even, and $-1$ for $i$ odd. The base cases are easy to check, especially if taking $i=0$ and $i=1$.
But first, we need an auxillary identity, which we will also use in $(A2)$.
\begin{equation}
P(n,i) = \sum_{j=0}^{i} (-1)^{i-j}\binom{n}{j} \binom{n-1-j}{i-j}
\end{equation}
We claim that $P(n,i)=1$ for all $n$ and $i$. This is certainly true for $i=0$. We split $\binom{n}{j}$ into two to get an induction:
\begin{align}
P(n,i) &= \sum_{j=0}^{i} (-1)^{i-j}\binom{n-1}{j} \binom{n-1-j}{i-j} + \sum_{j=1}^{i} (-1)^{i-j}\binom{n-1}{j-1} \binom{n-1-j}{i-j} \\
&= \sum_{j=0}^{i} (-1)^{i-j}\binom{n-1}{i} \binom{i}{j} + \sum_{j=1}^{i} (-1)^{i-j}\binom{n-1}{j-1} \binom{n-1-j}{i-j} \\
&= \binom{n-1}{i}\sum_{j=0}^{i} (-1)^{i-j}\binom{i}{j} + \sum_{j=0}^{i-1} (-1)^{i-1-j}\binom{n-1}{j} \binom{(n-1)-1-j}{(i-1)-j} \\
&= 0 + P(n-1,i-1)
\end{align}
Now we give a similar proof for $Q$. In the fourth to fifth lines, we used the alternating sum of binomial coefficients up to some number.
You might be able to give a direct proof if your generatingfunctionology is strong, since these look like convolutions of simple functions, but these proofs seemed easy enough that I didn't bother trying.
\begin{align}
Q(n,i) &= \sum_{j=0}^{i-1} \sum_{k=1}^{i-j} (-1)^k \binom{n}{k-1} \binom{n-k}{j} \\
&= \sum_{j=0}^{i-1} \sum_{k=1}^{i-j} (-1)^k \binom{n-1}{k-1} \binom{n-k}{j} + \sum_{j=0}^{i-2} \sum_{k=2}^{i-j} (-1)^k \binom{n-1}{k-2} \binom{n-k}{j} \\
&= \sum_{j=0}^{i-1} \sum_{k=1}^{i-j} (-1)^k \binom{n-1}{j} \binom{n-1-j}{k-1} + \sum_{j=0}^{i-2} \sum_{k=2}^{i-j} (-1)^k \binom{n-1}{k-2} \binom{n-k}{j} \\
&= \sum_{j=0}^{i-1} \binom{n-1}{j} \sum_{k=0}^{i-1-j} (-1)^{k+1} \binom{n-1-j}{k} + \sum_{j=0}^{i-2} \sum_{k=1}^{i-1-j} (-1)^{k+1} \binom{n-1}{k-1} \binom{n-1-k}{j} \\
&= \sum_{j=0}^{i-1} \binom{n-1}{j} (-1)^{i-1-j+1} \binom{n-2-j}{i-1-j} - Q(n-1, i-1) \\
&= -P(n-1,i-1) - Q(n-1, i-1)
\end{align}
Now, since our equation $(A1)$ was just $(n+1)Q(n,i)$, and $i$ is even, it's $0$. A similar treatment yields $(A2)$, again using the alternating binomial sum identity near the end:
\begin{align}
&\,\sum_{k=0}^i (-1)^k \binom{n+1}{k} \sum_{j=0}^{i-k} \binom{n-k}{j} \\
&= \sum_{j=0}^i \sum_{k=0}^{i-j} (-1)^k \binom{n+1}{k} \binom{n-k}{j} \\
&= \sum_{j=0}^i \sum_{k=0}^{i-j} (-1)^k \binom{n}{k} \binom{n-k}{j} + \sum_{j=0}^i \sum_{k=0}^{i-j} (-1)^k \binom{n}{k-1} \binom{n-k}{j} \\
&= \sum_{j=0}^i \sum_{k=0}^{i-j} (-1)^k \binom{n}{j} \binom{n-j}{k} + Q(n,i) \\
&= \sum_{j=0}^i \binom{n}{j} \sum_{k=0}^{i-j} (-1)^k \binom{n-j}{k} \\
&= \sum_{j=0}^i \binom{n}{j} (-1)^{i-j} \binom{n-1-j}{i-j} \\
&= P(n,i)
\end{align}
So we have shown that $(A) = i\cdot2^{i-2}$ for $i$ even. Let's use this for $(B)$, since $i-2$ is also even, we have:
\begin{equation}
\sum_{j=0}^{i-2} \binom{n-1-j}{i-2-j} 2^{j-1} \sum_{k=0}^j (-1)^k \binom{n+1}{k} 2^{i-k-3}(i-2-k) = (i-2)2^{i-4}
\end{equation}
After multiplying both sides by $2^2$, the only place the left side differs from $(B)$ is the extra $-2$ in $(i-2-k)$, and the right side is $2^{i-1}$ smaller than we would like.
So we show that
\begin{equation}
\sum_{j=0}^{i-2} \binom{n-1-j}{i-2-j} 2^{j-1} \sum_{k=0}^j (-1)^k \binom{n+1}{k} 2^{i-k} = 2^{i-1}
\end{equation}
Dividing out the factor of $2^{i-1}$ and switching the order of summation, we get
\begin{equation}
\sum_{k=0}^{i-2} (-1)^k \binom{n+1}{k} \sum_{j=k}^{i-2} \binom{n-1-j}{i-2-j}2^{j-k}
\end{equation}
Of course we recognize our hockey stick identity from earlier, so this simplifies to the case of $(A2)$
\begin{equation}
\sum_{k=0}^{i-2} (-1)^k \binom{n+1}{k} \sum_{j=0}^{i-2-k} \binom{n-k}{j} =1
\end{equation}
Best Answer
It isn't true. Choose $n,a,k$ such that $n+a-k>0$, $n+a-2k<0$ and $k>1$. The sum has only one nonzero term which is not equal to the right side.