Generalizing Catalan numbers: number of ways where we cross the diagonal $k$ times.

Instead of allowing possibly empty sections, we split a path at points of crossing the diagonal, and ignore possible touch points (where the path goes from above/below and bounces back). This gives $$R_{0,n}=2C_n,\qquad R_{k,n}=\sum_{m=1}^{n-1}C_m R_{k-1,n-m}\qquad(k,n>0)$$ ($R_{0,n}$ counts "Catalan" $n$-paths above/below the diagonal (not strictly); to get an $(k,n)$-path, we take an $(k-1,n-m)$-path and append a "Catalan" $m$-path which extends the last step). Then, in the notation of your question, $R_{k}(z):=\sum_{n=1}^{\infty}R_{k,n}z^n$ equals $2\big(B_2(z)-1\big)^{k+1}=2z^{k+1}B_2(z)^{2k+2}$. Using the identity $(5.70)$ from the question, we get $$R_{k-1}(z)=2z^k\sum_{t=0}^{\infty}\binom{2t+2k}{t}\frac{2k}{2t+2k}z^t\underset{n:=t+k}{\quad=\quad}2\sum_{n=k}^{\infty}\binom{2n}{n-k}\frac{k}{n}z^n,$$ that is, $R_{k-1,n}=\frac{2k}{n}\binom{2n}{n-k}$ for $1\leqslant k\leqslant n$.