Proof of (complicated?) summation equality

Consider the sum $$ S^n_m=\sum_{k=1}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}};\quad n\ge m\ge1.\tag{1} $$ We are going to prove: $$ S^n_m=\frac{2m}{2n-2m+1}.\tag{2} $$ It is easy to check that for $m=1$ and arbitrary $n\ge1$ the expression (2) is valid: $$ S^{n}_1=4^1\frac{\binom{n}{1}\binom{1}{1}}{\binom{2n}{2}\binom{2}{1}}=\frac{2}{2n-1}. $$

Assume that expression (2) is valid for some $S^{n-1}_{m-1}$ $(n\ge m\ge 2)$. It follows then that it is valid for $S^{n}_{m}$ as well: $$\begin {align} S^{n}_{m}&=\sum_{k=1}^{m}4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\ &=\sum_{k=1}^{m}4^k\frac{\frac{n}{k}\binom{n-1}{k-1}\frac{m}{k}\binom{m-1}{k-1}}{\frac{2n(2n-1)}{2k(2k-1)}\binom{2n-2}{2k-2}\frac{2k(2k-1)}{k^2}\binom{2k-2}{k-1}}\\ &=\frac{2m}{2n-1}\sum_{k=1}^{m}4^{k-1}\frac{\binom{n-1}{k-1}\binom{m-1}{k-1}}{\binom{2n-2}{2k-2}\binom{2k-2}{k-1}}\\ &=\frac{2m}{2n-1}\sum_{k=0}^{m-1}4^{k}\frac{\binom{n-1}{k}\binom{m-1}{k}}{\binom{2n-2}{2k}\binom{2k}{k}}\\ & =\frac{2m}{2n-1}\left(S^{n-1}_{m-1}+1\right)\\ &\stackrel{I.H.}{=}\frac{2m}{2n-1}\left(\frac{2(m-1)}{2(n-1)-2(m-1)+1}+1\right)\\ & =\frac{2m}{2n-1}\cdot\frac{2n-1}{2n-2m+1}\\ &=\frac{2m}{2n-2m+1},\end {align} $$ where $\stackrel{I.H.}{=}$ means substitution of the induction assumption. Thus, the equality (2) is proved.

The rest is rather straightforward. Let $l$ be an integer number $(1\le l\le n)$. Then: $$\begin {align} \sum_{m=1}^l\frac{2}{2n-2m+1}&\stackrel {(2)}=\sum_{m=1}^l\frac{1}{m}\sum_{k=1}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\ &= \sum_{m=1}^l\frac{1}{m}\sum_{k=1}^\infty4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\ &=\sum_{k=1}^\infty4^k\frac{\binom{n}{k}}{\binom{2n}{2k}\binom{2k}{k}}\sum_{m=1}^l\frac{1}{m}\binom{m}{k}\\ &\stackrel{!}{=}\sum_{k=1}^\infty\frac{4^k}{k}\frac{\binom{n}{k}\binom{l}{k}}{\binom{2n}{2k}\binom{2k}{k}}\\ &=\sum_{k=1}^l\frac{4^k}{k}\frac{\binom{n}{k}\binom{l}{k}}{\binom{2n}{2k}\binom{2k}{k}}.\tag{3}\end {align} $$

The proof of identity: $$ \sum_{m=1}^l\frac{1}{m}\binom{m}{k}=\frac{1}{k}\binom{l}{k};\quad k\ge1, $$ used in $\stackrel{!}{=}$, is left as an exercise.

The equality (3) is identical to that one claimed in question. To see it redefine in the latter $N=2n$, $h=2(n-l)$ and reverse the summation order in RHS: $j\mapsto l+1-j$.

To complement the nice answer provided by user355705, please refer to the answer provided in this related post where it is shown that $$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{0\, \le \,k\, \le \,m} {4^{\,k} {{\left( \matrix{ n \cr k \cr} \right)\left( \matrix{ m \cr k \cr} \right)} \over {\left( \matrix{ 2n \cr 2k \cr} \right)\left( \matrix{ 2k \cr k \cr} \right)}}} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {{{\left( \matrix{ m \cr k \cr} \right)} \over {\left( \matrix{ n - 1/2 \cr k \cr} \right)}}} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ m \cr k \cr} \right)1^{\,\overline {\,k\,} } \left( {n + 1/2} \right)^{\,\overline {\, - \,k\,} } } \cr & = {1 \over {\left( {n - m + 1/2} \right)^{\,\overline {\,m\,} } }}\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\left( \matrix{ m \cr k \cr} \right)1^{\,\overline {\,k\,} } \left( {n - m + 1/2} \right)^{\,\overline {\,m - \,k\,} } } = \cr & = {{\left( {n - m + 3/2} \right)^{\,\overline {\,m\,} } } \over {\left( {n - m + 1/2} \right)^{\,\overline {\,m\,} } }} = {{n + 1/2} \over {n - m + 1/2}} = {{2n + 1} \over {2(n - m) + 1}} \cr} } $$

This identity is valid for any non-negative integer $m$, and for $n$ that can take even a real or complex value, except for $n=m-1/2$.
This under the acception that the two binomials in $n$, when null get simplified. That is that the two binomials be rewritten as above in terms of Falling Factorials (or Gamma function) with $n$ real, and simplify the fraction (take the limit).