Alternatively, we can prove this along the lines of the reflection method used for the standard Catalan numbers. ([Read about][1] this first if you have never seen it.)

There are $\binom{2n}{n}$ paths from the lower hand corner (0,0) to the upper right hand corner (n,n) of the square. 

[![enter image description here][2]][2]

For the paths that move above the diagonal, we reflect after the first point above the diagonal (a), over the line through (a) and (n,n), to find a bijection between paths that move above the diagonal and paths from (0,0) to (n-1,n+1).

[![enter image description here][3]][3][![enter image description here][4]][4]

This gives us the usual Catalan numbers: $$C_n=\binom{2n}{n}-\binom{2n}{n-1}.$$

We can count paths that cross the line through (k,0) and $(n,k)$ in the same manner. We have $$\binom{2n}{n-k-1}$$ such paths. 

[![enter image description here][5]][5][![enter image description here][6]][6]

This will help us find the paths of height at most $k$, but we can't simply subtract this number from the Catalan numbers, because we must take into account paths that cross both lines. This can happen in 3 ways (when $\frac{n-1}{2} \leq k \leq n$). 

 - A path can cross the diagonal and then cross the $k$-line. Using a double-reflection bijection we see that there are $$\binom{2n}{n-k-2}$$ such paths.

[![enter image description here][7]][7][![enter image description here][8]][8]

 - A path can cross the $k$-line and then the diagonal. Using a double-reflection bijection we see that there are $$\binom{2n}{n-k-2}$$ such paths.

[![enter image description here][9]][9][![enter image description here][10]][10]

 - The above two sets of paths intersect at the third option: our path crosses the diagonal, then the $k$-line, and then the diagonal again. We do a triple-reflection bijection to see that there are $$\binom{2n}{n-k-3}$$ such paths.

[![enter image description here][11]][11][![enter image description here][12]][12]

Putting all of this together, we find the number of Dijck paths of height at most $k$, $\frac{n-1}{2} \leq k \leq n$, as $$S(n,k)=\binom{2n}{n}-\binom{2n}{n{-}1}-\binom{2n}{n{-}k{-}1}+2\binom{2n}{n{-}k{-}2}-\binom{2n}{n{-}k{-}3}.$$

Therefore, the number of paths of height exactly $k$, when $\frac{n+1}{2} \leq k \leq n$, is $$T(n,k)=S(n,k)-S(n,k{-}1)=\binom{2n}{n{-}k}-3\binom{2n}{n{-}k{-}1}+3\binom{2n}{n{-}k{-}2}-\binom{2n}{n{-}k{-}3}.$$

(I used Wolfram to verify that this is indeed the $T(n,k)$ that you mention.)


  [1]: https://www.isical.ac.in/~sush/Discrete-maths-2014/Catalan-numbers.pdf
  [2]: https://i.sstatic.net/WFFTf.png
  [3]: https://i.sstatic.net/CAhGV.png
  [4]: https://i.sstatic.net/q7gf9.png
  [5]: https://i.sstatic.net/0uMqj.png
  [6]: https://i.sstatic.net/w6XuH.png
  [7]: https://i.sstatic.net/CBvac.png
  [8]: https://i.sstatic.net/2h67B.png
  [9]: https://i.sstatic.net/vAI4e.png
  [10]: https://i.sstatic.net/EEI3h.png
  [11]: https://i.sstatic.net/XxuNK.png
  [12]: https://i.sstatic.net/R989w.png