Your question sounds very similar to
this one,
and my answer is similar as well: de Bruijn's book *Asymptotic Methods in Analysis* discusses
this type of problems in great detail.

Write your sum as $S_k=\sum_{j=1}^ka_j$ where, as you mention,
$$
\frac{a_{j+1}}{a_j}
=\frac14\biggl(\frac{k-1}{j-1}-1\biggr)\biggl(\frac{n-k-1}{j-1}-1\biggr)
$$
monotonically decreases (as the function of $j$) from $\infty$ to $0$.
This means that $a_j$ first increases for $j\le m$ and then decreases
for $j\ge m$, where $m=m(k,n)$ is determined by $a_{m+1}/a_m\approx 1$.
In the your example, $m-1$ is a solution of the quadratic equation. Then
the main term of the asymptotics is fully determined by the growth of $a_m$
in the sense
$$
\lim_{k\to\infty}S_k^{1/k}
=\lim_{k\to\infty}a_{m(k,n)}^{1/k},
$$
and then you can use Stirling's formula to determine the latter limit explicitly.
(de Bruijn's book also contains details on how to compute more accurate
asymptotics.)

Alternatively, you can use Zeilberger's algorithm of creative telescoping
to find a recurrence equation for $S_k$; then the same asymptotics can be
read from its characteristic equation.