Let $$B(z)=\frac{1-\sqrt{1-4z}}{2z}$$ and $$T(z)=\frac{1-\sqrt{1-4z^6}}{2z^2}$$
I know that $[z^n]B(z)=\frac{1}{n+1}\binom{2n}{n}$ (Catalan numbers).
We have that $T(z)=z^4B(z^6)$, so $[z^n]T(z)=[z^{n-4}]B(z^6)$
I thought this would imply that $[z^n]T(z)=\frac{1}{k+1}\binom{2k}{k}$ when $n-4=6k$ but the correct solution is $\frac{1}{4k-2}\binom{2k}{k}$ when $n=6k-2$.
Where is my error?
Both answers are right, because both are the same.
Since your attempt writes $n=6k+4=6(k+1)-2$ whereas the "correct solution" you've looked up somewhere writes $n=6k-2$, the $k$s are different. Accordingly, rewrite your inferred answer as $\frac1k\binom{2k-2}{k-1}$ for easier comparison with the desired result. Finally, note$$\frac{\frac1k\binom{2k-2}{k-1}}{\frac{1}{4k-2}\binom{2k}{k}}=\frac{(2k-2)!(4k-2)k!^2}{(k-1)!^2k(2k)!}=\frac{(4k-2)k^2}{k(2k-1)2k}=1.$$
