There is a pleasant, natural statistical solution to this problem for integral values of $K$, showing that the product has a $\chi^2(2)$ distribution. It relies only on well-known, easily established relationships among functions of standard normal variables.
When $K$ is integral, a Beta$(1,K-1)$ distribution arises as the ratio $$\frac{X}{X+Z}$$ where $X$ and $Z$ are independent, $X$ has a $\chi^2(2)$ distribution, and $Z$ has a $\chi^2(2K-2)$ distribution. (See the Wikipedia article on the Beta distribution for instance.)
Any $\chi^2(n)$ distribution is that of the sum of squares of $n$ independent standard Normal variates. Consequently, $X+Z$ is distributed as the squared length of a $2 + 2K-2 = 2K$ vector with a standard multinormal distribution in $\mathbb{R}^{2K}$ and $X/(X+Z)$ is the squared length of the first two components when that vector is radially projected to the unit sphere $S^{2K-1}$.
The projection of a standard multinormal $n$-vector onto the unit sphere has a uniform distribution because the multinormal distribution is spherically symmetric. (That is, it is invariant under the orthogonal group, a result that follows immediately from two simple facts: (a), the orthogonal group fixes the origin and by definition does not change covariances; and (b) the mean and covariance completely determine the multivariate normal distribution. I illustrated this for the case $n=3$ at https://stats.stackexchange.com/a/7984). In fact, the spherical symmetry immediately shows this distribution is uniform conditional on the length of the original vector. The ratio $X/(X+Z)$ therefore is independent of the length.
What all this implies is that multiplying $X/(X+Z)$ by an independent $\chi^2(2K)$ variable $Y$ creates a variable with the same distribution as $X/(X+Z)$ multiplied by $X+Z$; to wit, the distribution of $X$, which has a $\chi^2(2)$ distribution.