This identity is known as Jonah's formula (special case with $n\rightarrow 2n$ and $r\rightarrow n$, see "Catalan Numbers with Applications" by Thomas Koshy, pg. 325-326 for a combinatorial proof)
$$\sum_{k=0}^r\binom{n-2k}{r-k}C_k=\binom{n+1}r$$
and a $q$-analogue was obtained by Andrews in "$q$-Catalan identities" in the book "The legacy of Alladi Ramakrishnan in the Mathematical Sciences". It's Theorem 3, pg. 186.
$$\frac{(1+q^{n-r+1})}{(1+q^{r+1})}\left[ {\begin{array}{c}n+1\\r\end{array} } \right]_{q^2}=-(-q\;;q)_{n+1}\sum_{k=0}^r\left[ {\begin{array}{c}n-2k\\r-k\end{array} } \right]_{q^2}\frac{\textrm{C}_{k+1}(-1;q)}{(-q\;;q)_{n-2k}}q^{-k-1}$$
where $\textrm{C}_n(\lambda,q)$ is a $q$-analogue of the Catalan numbers considered also by Andrews here.
$$\textrm{C}_n(\lambda,q)=\frac{q^{2n}(-\lambda/q; q^{2})_{n}}{(q^2;q^2)_{n}}$$
In the paper, he says that the general strategy is to go from a binomial coefficient identity to a generalized hypergeometric identity, and then we can look for a $q$-analogue of the latter. In this case, he used the Pfaff-Saalschütz summation formula and then he searched for a $q$-analogue of this one with the help of Bailey's and Gasper and Rahman's books. I can't help much more, I'm not familiar with these kind of hypergeometric identities.
If $n\rightarrow 2n$ and $r\rightarrow n$, the limit $q\rightarrow 1$ recovers the identity (1).
Best Answer
By the ballot theorem, $\frac{k}{n} \binom{2n}{n+k}$ is the number of Dyck paths, i.e. $(1,1), (1,-1)$-walks in the quadrant, from the origin to $(2n-1, 2k-1)$. You need to concatenate a pair of those to get a Dyck path to $(4n-2,0)$, and $k$ takes values between 1 and $n$.