We can rewrite the sum as
$$
\eqalign{
& S(k) = {1 \over {2^{\,2k} }}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
\left( \matrix{ 2k \cr m \cr} \right)\left( \matrix{ 4k - 2m \cr 2k \cr} \right)
{1 \over {2k - 2m + 1}}} = \cr
& = {1 \over {2^{\,2k} }}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
\left( \matrix{ 2k \cr m \cr} \right){{\left( {4k - 2m} \right)^{\,\underline {\,2k\,} } }
\over {\left( {2k} \right)!}}{1 \over {\left( {2k - 2m + 1} \right)}}} = \cr
& = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
\left( \matrix{ 2k \cr m \cr} \right)
\left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} } } \cr}
$$
Now the Falling factorial
$$
p(m,k) = \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} }
$$
is a polynomial in $m$ with the following characteristics.
$$
\left\{ {\matrix{
{k = 0\quad \Rightarrow \quad } & \matrix{
p(m,0) = \left( { - 2m} \right)^{\,\underline {\, - 1\,} }
= {1 \over {\left( { - 2m + 1} \right)^{\underline {\,1\,} } }}
= {1 \over { - 2m + 1}}\quad \Rightarrow \hfill \cr
\Rightarrow \quad p(0,0) = 1 \hfill \cr} \cr
{1 \le k\quad \Rightarrow \quad } & \matrix{
p(m,k) = \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} } \hfill \cr
{\rm polynomial}\,{\rm in}\,{\rm m}\,{\rm ofdegree}\;2k - 1 \hfill \cr
{\rm with}\,{\rm zeros} \in \left[ {k + 1,\;2k} \right] \hfill \cr} \cr
} } \right.
$$
Therefore we have
$$
S(0) = 1
$$
and for $1 \le k$
$$
S(k)\quad \left| {\;1 \le k} \right. = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^k
{\left( { - 1} \right)^{\,m} \left( \matrix{ 2k \cr m \cr} \right)p(m,k)}
$$
Since $p(m,k) = 0$ for $k+1 \le m \le 2m$ we can extend the sum to $2k$ and multiply the summand by $1 = (-1)^{2k}$
$$
\eqalign{
& S(k)\quad \left| {\;1 \le k} \right.\quad
= {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^{2k}
{\left( { - 1} \right)^{\,m} \left( \matrix{ 2k \cr m \cr} \right)p(m,k)} = \cr
& = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^{2k}
{\left( { - 1} \right)^{2k - \,m} \left( \matrix{ 2k \cr m \cr} \right)p(m,k)} \cr}
$$
Here we recognize that the sum represents the finite difference (unitary step)
of order $2k$ of $p(m,k)$, and it is known (re. to the Newton series)
that the difference of a polynomial , of order greater than the degree of the same is identically null.
In conclusion $$S(k)= \delta _{\,k, \, 0} =\binom {0}{k}$$