By reflecting paths over the diagonal, one can see that the number of paths that never go below the diagonal equals the number of paths that never go above the diagonal.
However, it's possible for a path to be below the diagonal at one point and above the diagonal at another point, so the total number of paths is greater than twice the Catalan number.
I find it a bit easier to think in terms of paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$ that consist of $n$ up-steps (steps from $\langle k,\ell\rangle$ to $\langle k+1,\ell+1\rangle$) and $n$ down-steps (steps from $\langle k,\ell\rangle$ to $\langle k+1,\ell-1\rangle$). An up-step in this version corresponds to a step to the right in your version, and a down-step corresponds to a step upwards in your version. Your boundary condition becomes a requirement that my path not drop below the line $y=-z$.
We can use a slight modification of one of the usual arguments for counting the paths that don’t drop below the line $y=0$.
As in your version, there are altogether $\binom{2n}n$ paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$, and the problem is to count the ‘bad’ ones, i.e., the ones that do drop below the line $y=-z$. Suppose that we have a bad path $\pi$. There is a first point at which $\pi$ reaches the line $y=-z-1$; if it has made $u$ up-steps at that point, it must have made $u+z+1$ down-steps and so have reached the point $\langle 2u+z+1,-z-1\rangle$. Reflect the remainder of $\pi$ (i.e., the part to the right of this point) in the line $y=-z-1$. That part of $\pi$ has $n-u$ up-steps and $n-u-z-1$ down-steps, so its reflection has $n-u$ down-steps and $n-u-z-1$ up-steps. This means that it must end at the point
$$\langle 2u+z+1,-z-1\rangle+\langle2n-2u-z-1,-z-1\rangle=\langle 2n,-2z-2\rangle\;.$$
Conversely, any path from $\langle 0,0\rangle$ to $\langle 2n,-2z-2\rangle$ must hit the line $y=-z-1$, and if we reflect the part of it to the right of that intersection in the line $y=-z-1$, we get a path from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$ that drops below the line $y=-z$. Thus, we have a bijection between our bad paths and all paths from $\langle 0,0\rangle$ to $\langle 2n,-2z-2\rangle$. Each of these paths has $n-z-1$ up-steps and $n+z+1$ down-steps, so there are $\binom{2n}{n+z+1}$ of them. Thus, there are
$$\binom{2n}n-\binom{2n}{n+z+1}=\binom{2n}n-\binom{2n}{n-z-1}$$
good paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$.
Best Answer
I find it easier to work with up-down paths, defined as sequence of up-steps (e.g., from $\langle m,n\rangle$ to $\langle m+1,n+1\rangle$) and downsteps (e.g., from $\langle m,n\rangle$ to $\langle m+1,n-1\rangle$). A Dyck path is then an up-down path that does not go below the $x$-axis, and there are $C_n$ Dyck paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$.
If $n,n'\ge 0$, $m<m'$, and $t\ge 0$, let $\mathscr{D}(m,n,m',n',t)$ be the set of Dyck paths from $\langle m,n\rangle$ to $\langle m',n'\rangle$ that hit the $x$-axis exactly $t$ times, and let $d(m,n,m',n',t)=|\mathscr{D}(m,n,m',n',t)|$. Thus, $d(0,0,2n,0,2)=C_n$.
Suppose that $t\ge 1$, and $0\le k<t$. Let $P\in\mathscr{D}(m,n,m',n',t)$. Let the points where $P$ hits the $x$-axis be $\langle\ell_i,0\rangle$ for $i=1,\ldots,t$, where $\ell_1<\ldots<\ell_t$. Since $P$ is a Dyck path, $P$ has an up-step immediately after each of these points except possibly the last. Remove the first $k$ of these up-steps, and translate the resulting path upwards $k$ units to get a path $\hat P$; it’s straightforward to check that $\hat P\in\mathscr{D}(m,n+k,m'-k,n',t-k)$: the path has been shortened by $k$ steps, and the first $k$ hits have been eliminated.
This procedure is reversible. If $P\in\mathscr{D}(m,n+k,m'-k,n',t-k)$, we first shift $P$ down by $k$ units to get an up-down path $P'$. For $i=1,\ldots,k$ let $\ell_i$ be minimal such that $\langle\ell_i,1-i\rangle$ is on $P'$. Let $Q$ be the up-down path obtained from $P'$ by inserting an up-step immediately after each of the points $\langle\ell_i,1-i\rangle$. Then $Q\in\mathscr{D}(m,n,m',n',t)$, and $Q=\hat P$. It follows that
$$d(m,n,m',n',t)=d(m,n+k,m'-k,n',t-k)\;.$$
In particular,
$$d(0,0,2n,0,k+1)=d(0,k,2n-k,0,1)=d(0,0,2n-k,k,1)\;.$$
Each path in $\mathscr{D}(0,0,2n-k,k,1)$ must begin with an up-step; if we remove this up-step and shift the path one unit down and to the left, we get a Dyck path from $\langle 0,0\rangle$ to $\langle 2n-k-1,k-1\rangle$. Conversely, any Dyck path from $\langle 0,0\rangle$ to $\langle 2n-k-1,k-1\rangle$ becomes a path in $\mathscr{D}(0,0,2n-k,k,1)$ if we shift it one unit up and to the right and prepose an up-step. Thus,
$$d(0,0,2n-k,k,1)=\sum_{t\ge 1}d(0,0,2n-k-1,k-1,t)\;.$$
Each up-down path from $\langle 0,0\rangle$ to $\langle 2n-k-1,k-1\rangle$ corresponds to a possible sequence of $2n-k-1$ votes, $n-1$ of them for candidate $A$ and $n-k$ of them for candidate $B$. On this interpretation the Dyck paths in $\mathscr{D}(0,0,2n-k-1,k-1,1)$ correspond to the sequences in which candidate $A$ is never behind candidate $B$. Calculating the number of such paths is essentially the ballot problem.
Clearly there are $\binom{2n-k-1}{n-1}$ up-down paths from $\langle 0,0\rangle$ to $\langle 2n-k-1,k-1\rangle$. If $P$ is one of these paths that is not a Dyck path, there is a least $\ell$ such that $\langle\ell,-1\rangle$ is on the path $P$. Reflect the initial segment of $P$ between $\langle 0,0\rangle$ and $\langle\ell,-1\rangle$ in the line $y=-1$; the result is a path $P'$ from $\langle 0,-2\rangle$ to $\langle 2n-k-1,k-1\rangle$ that first hits the line $y=-1$ at $x=\ell$. Note that if this initial segment has $u$ up-steps, then it has $u+1$ down-steps, and the rest of $P$ has $n-1-u$ up-steps. As a result, $P'$ has $(n-1-u)+(u+1)=n$ up-steps and $n-k-1$ down-steps.
Conversely, if $P$ is an up-down path from $\langle 0,-2\rangle$ to $\langle 2n-k-1,k-1\rangle$ that first hits the line $y=-1$ at $x=\ell$, reflecting the first $\ell$ steps of $P$ in the line $y=-1$ produces an up-down path $Q$ from $\langle 0,0\rangle$ to $\langle 2n-k-1,k-1\rangle$ that first hits the line $y=-1$ at $x=\ell$, and it’s clear that $Q'=P$. Every up-down path from $\langle 0,-2\rangle$ to $\langle 2n-k-1,k-1\rangle$ hits the line $y=-1$, and there are $\binom{2n-k-1}{n}$ such paths (since each has $n$ up-steps and $n-k-1$ down-steps). Thus, the number of up-down paths from $\langle 0,0\rangle$ to $\langle 2n-k-1,k-1\rangle$ that are not Dyck paths is $\binom{2n-k-1}n$, and
$$\begin{align*} d(0,0,2n,0,k+1)&=\binom{2n-k-1}{n-1}-\binom{2n-k-1}n\\ &=\binom{2n-k-1}{n-k}-\binom{2n-k-1}{n-k-1}\\ &=\binom{2n-k-1}{n-k}\left(1-\frac{n-k}{n}\right)\\ &=\binom{2n-k-1}{n-k}\frac{k}n\;. \end{align*}$$
The Dyck paths in $\mathscr{D}(0,0,2n,0,k+1)$ correspond naturally to the paths in the plane from $\langle 0,0\rangle$ to $\langle n,n\rangle$ using only right-steps and up-steps that hit the diagonal exactly $k-1$ times between the two endpoints, or $k+1$ times including the endpoints.