Let's refer everything to square matrices indexed from $0$ to $h$, that I will denote as
$$
{\bf M}_{\,h}  = \left\| {\;f(n,m)\;} \right\|_{\,h} 
$$
with $n$ being the row index and $m$ the column index.

I will then denote by
$$
\left( {f(n) \circ {\bf I}_{\,h} } \right)
$$
the diagonal matrix whose entries are equal to $f(n)$.

So I write  the matrix you proposed as
$$ \bbox[lightyellow] {  
{\bf M}_{\,h} (a) = \left\| {\;4^{ - n - a} \left( \matrix{
  2n + 2a \cr 
  m + a \cr}  \right)\;} \right\|_{\,h}  = \left( {4^{ - n - a}  \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{
  2n + 2a \cr 
  m + a \cr}  \right)\;} \right\|_{\,h} 
 }$$
where $h=b-a$.

That premised, consider that in general
$$
\eqalign{
  & \left( \matrix{
  r\,n + t \cr 
  m + q \cr}  \right) = {{\left( {r\,n + t} \right)^{\,\underline {\,m + q} } } \over {\left( {m + q} \right)!}} = {{\left( {r\,n + t} \right)^{\,\underline {\,q} } \left( {r\,n + t - q} \right)^{\,\underline {\,m} } } \over {\left( {m + q} \right)^{\,\underline {\,q} } \;m^{\,\underline {\,m} } }} =   \cr 
  &  = \left( {r\,n + t} \right)^{\,\underline {\,q} } \left( \matrix{
  r\,n + t - q \cr 
  m \cr}  \right){1 \over {\left( {m + q} \right)^{\,\underline {\,q} } }} \cr} 
$$
where $x^{\,\underline {\,a} } $ denotes the falling factorial ($x^{\overline {\,a\,} } $ the rising) and
where, for the present problem, we consider $q$ to be a non-negative integer,
while $r$ and $t$ could be real (or even complex).

Then we have that we can write the binomial as
$$
\begin{gathered}
  \left( \begin{gathered}
  r\,n + s \\ 
  m \\ 
\end{gathered}  \right) = \frac{1}
{{m!}}\left( {r\,n + s} \right)^{\,\underline {\,m\,} }  = \frac{1}
{{m!}}\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right)} {\left( \begin{gathered}
  m \\ 
  k \\ 
\end{gathered}  \right)s^{\,\underline {\,m - k\,} } \left( {r\,n} \right)^{\,\underline {\,k\,} } }  =  \hfill \\
   = \frac{1}
{{m!}}\sum\limits_{\left\{ \begin{subarray}{l} 
  \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right) \\ 
  \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \;h} \right) 
\end{subarray}  \right.} {\left( \begin{gathered}
  m \\ 
  k \\ 
\end{gathered}  \right)s^{\,\underline {\,m - k\,} } \left( { - 1} \right)^{\,k - j} \left[ \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right]r^{\,j} n^{\,j} }  =  \hfill \\
   = \sum\limits_{\left\{ \begin{subarray}{l} 
  \left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \;h} \right) \\ 
  \left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \;h} \right) 
\end{subarray}  \right.} {n^{\,j} r^{\,j} \left( { - 1} \right)^{\,k - j} \left[ \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right]\frac{1}
{{k!}}\left( \begin{gathered}
  s \\ 
  m - k \\ 
\end{gathered}  \right)}  \hfill \\ 
\end{gathered} 
$$

Then in the last line we can replace $n^m$ with
$$
n^{\,m}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \;h} \right)} {\left\{ \matrix{
  m \cr 
  k \cr}  \right\}n^{\,\underline {\,k\,} } }  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \;h} \right)} {\left( \matrix{
  n \cr 
  k \cr}  \right)k!\left\{ \matrix{
  m \cr 
  k \cr}  \right\}} 
$$

Thus we arrive finally to
$$ \bbox[lightyellow] {  
\eqalign{
  & {\bf M}_{\,h} (a) = \left\| {\;4^{ - n - a} \left( \matrix{
  2n + 2a \cr 
  m + a \cr}  \right)\;} \right\|_{\,h}  = \left( {4^{ - n - a}  \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{
  2n + 2a \cr 
  m + a \cr}  \right)\;} \right\|_{\,h}  =   \cr 
  &  = \left( {4^{ - n - a}  \circ {\bf I}_{\,h} } \right)\left( {\left( {2\,n + 2a} \right)^{\,\underline {\,a} }  \circ {\bf I}_{\,h} } \right)\;\left\| {\;\left( \matrix{
  2n + a \cr 
  m \cr}  \right)\;} \right\|_{\,h} \left( {{1 \over {\left( {n + a} \right)^{\,\underline {\,a} } }} \circ {\bf I}_{\,h} } \right) =   \cr 
  &  = \left( {4^{ - n - a}  \circ {\bf I}_{\,h} } \right)\left( {\left( {2\,n + 2a} \right)^{\,\underline {\,a} }  \circ {\bf I}_{\,h} } \right)\;{\bf B}_{\,h} \left( {n! \circ {\bf I}_{\,h} } \right)\;\overline {{\bf St}_{{\bf 2}\,h} } \left( {2^{\,n}  \circ {\bf I}_{\,h} } \right)\;\overline {{\bf St}_{{\bf 2}\,h} } ^{\,{\bf  - }\,{\bf 1}} \left( {n! \circ {\bf I}_{\,h} } \right)^{\,{\bf  - }\,{\bf 1}} \left( {{\bf I}_{\,h}  + \overline {{\bf E}_{\,h} } } \right)^{\,{\bf a}} \left( {{1 \over {\left( {n + a} \right)^{\,\underline {\,a} } }} \circ {\bf I}_{\,h} } \right) \cr} 
 }$$
with
$$
\eqalign{
  & {\bf B}_{\,h}  = \;\left\| {\;\left( \matrix{
  n \cr 
  m \cr}  \right)\;} \right\|_{\,h} \quad {\bf St}_{{\bf 2}\,h}  = \;\left\| {\;\left\{ \matrix{
  n \cr 
  m \cr}  \right\}\;} \right\|_{\,h} \quad {\bf I}_{\,h}  + {\bf E}_{\,h}  = \;\left\| {\;\left( \matrix{
  1 \cr 
  n - m \cr}  \right)\;} \right\|_{\,h}   \cr 
  & \overline {\bf X}  = transpose({\bf X}) \cr} 
$$

After that the determinant follows easily, since the matrices other than the diagonal ones
have unitary determinant
$$ \bbox[lightyellow] {  
\left| {\,{\bf M}_{\,h} (a)\,} \right| = \left( {\prod\limits_{0\, \le \,n\, \le \;h} {{{\left( {2\,\left( {n + a} \right)} \right)^{\,\underline {\,a} } } \over {2^{\,n + 2a} \left( {n + a} \right)^{\,\underline {\,a} } }}} } \right) = \left( {\prod\limits_{0\, \le \,n\, \le \;h} {{{\left( \matrix{
  2\,\left( {n + a} \right) \cr 
  a \cr}  \right)} \over {2^{\,n + 2a} \left( \matrix{
  n + a \cr 
  a \cr}  \right)}}} } \right)
 }$$