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,
$$d(0,0,2n,0,k+1)=d(0,k,2n-k,0,1)=d(0,0,2n-k,k,1)\;.$$
This bijection between paths $P$ and $\hat P$ is the up-down version of the correspondence that is very, very badly described at your CodeChef link, where its author writes ‘This mapping is same as decreasing value of $y$ for a path when it touches a diagonal’.
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.
Best Answer
We can count the number of paths from $(0,0)$ to $(2n+m,m)$, where each step is $(+1,+1)$ or $(+1,-1)$, and which never go below the $x$-axis, using the reflection principle.
There are $\binom{2n+m}{n}$ paths total, including the "bad" paths which touch the line $y=-1$ at some point. Given a bad path, $P$, define a new path, $P_\text{refl}$ as follows. Find the first point, where $P$ intersects the line $y=-1$, and vertically reflect every step after that point to get $P_{\text{refl}}$.
Since $P$ went from the line $y=-1$ up to the line $y=m$ at the end, the reflected path $P_\text{refl}$ will instead go from $y=-1$ down to $y=-m-2$. That is, $P_\text{refl}$ is a path from $(0,0)$ to $(2n+m,-m-2)$. Furthermore, you can show that this reflection map is a bijection from the set of bad paths to $(2n+m,m)$, to the set of all paths from $(0,0)$ to $(2n+m,-m-2)$. It follows that the number of bad paths is $$ \binom{2n+m}{n-1}. $$ Therefore, the number of good paths is $$ \binom{2n+m}{n}-\binom{2n+m}{n-1}. $$