Let
$$f(a,b,t) = \sqrt{1-at}\sqrt{1-bt}$$
We take the series for $\sqrt{1-at}$ and $\sqrt{1-bt}$ around $t = 0$ and multiply them together to find
$$f(a,b,t) = \sum_{n=0}^\infty \frac{t^n}{4^n}\sum_{k=0}^n \frac{(2k)!(2n-2k)!}{(1-2k)(1-2n+2k)k!^2 (n-k)!^2}a^k b^{n-k}\,.$$
Can we find some simpler representation of these coefficients? Obviously some sort of closed form would be nice. If we compute the first few terms we have
$$\begin{align*}f(a,b,t)= & \,\,1-\frac{1}{2}(a+b)t-\frac{1}{8}(a-b)^2t^2-\frac{1}{16}(a-b)^2(a+b)t^3\\&-\frac{1}{128}(a-b)^2(5a^2+6ab+5b^2)+\cdots\end{align*}$$
and that's what gave me the inspiration to ask this question. Now let
$$g(a,b,t) = \frac{-8}{(a-b)^2t^2}\left(f(a,b,t)-1+\frac{1}{2}(a+b)t\right)$$
we have the expansion
$$g(a,b,t) = 1+\frac{1}{2}(a+b)t+\frac{1}{16}(5a^2+6ab+5b^2)t^2\\\quad\quad\quad+\frac{1}{32}(7a^3+9a^2b+9ab^2+7b^3)t^3+\cdots$$
which seems to get at the more difficult coefficients. Thanks for your help!