If I'm not mistaken, at least for even $k$, this is solvable, but it is very tedious. Following is the outline, in which, for simplicity, I'm omitting constant terms from any integral (i.e., those not involved in the integration).
The ratio distribution of two independent $\chi^2$ RVs is
$$
\sim \int_y |y| (zy)^{\frac{k}{2} - 1}e^{-\frac{zy}{2}} y^{\frac{k}{2} - 1}e^{-\frac{y}{2}} \text{d}y
=
\int_y |y| (zy^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y
\; (1)
$$
(again, note that constants are omitted from the integral).
Similarly, if you shift the RVs by $a$ (which, in your case is the mean), then the ratio distribution is
$$
\sim \int_y |y| (zy - a)^{\frac{k}{2} - 1}e^{-\frac{zy}{2}} (y - a)^{\frac{k}{2} - 1}e^{-\frac{y}{2}} \text{d}y
\\
=
\int_y |y| (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y
\; (2)
$$
(once again, note that constants are omitted from the integral).
Note that
$$
\int_y |y| (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y \; (3)
\\
=
\int_{y > 0} y (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y
-
\int_{y < 0} y (zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1}e^{-\frac{y(z + 1)}{2}} \text{d}y
.
$$
If $k$ is even, then $q = \frac{k}{2} - 1$ is an integer, and, in (3),
$$
(zy^2 - ay(z + 1) + a^2)^{\frac{k}{2} - 1} = (zy^2 - ay(z + 1) + a^2)^q
$$
so that trinomial expansion can be used. The integral corresponding to each term in the expansion has a closed form solution.
I wish you the best of luck!!!