You can also prove this from the perspective of complex analysis. Let \(z\) and \(w\) be complex numbers such that \(z^{2n}=w\), and let \(z=r\exp\left(i\theta_0\right)\) so that \(|z|=r\) and \(\arg z=\theta_0\).
The \(2n\)th roots of \(w\) are then
\[\large w^{\frac{1}{2n}}=z=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\]
As you probably know, the \(n\)th roots of any complex number form a regular \(n\)-gon around the origin.
A reflection across the real axis (a "horizontal flip", as you put it) is equivalent to the mapping \(\xi\mapsto\bar{\xi}\) for any \(z\) in the plane, where \(\bar{\xi}\) is the complex conjugate of \(\xi\). A reflection across the imaginary axis ("vertical flip") is the same as \(\xi\mapsto-\bar{\xi}\). Putting these transformations together yields \(\xi\mapsto-\xi\). So, any of the \(2n\)th roots \(z\) simply become \(-z\), which can be written as
\[\large -z=-r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\]
On the other hand, a rotation of \(180^\circ\), or \(\pi\text{ rad}\), can be described by writing \(\arg z\) as \(\arg z^*:=\arg z+i\pi\), so we have
\[\large z^*=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2n\pi+2\pi k}{2n}i\right),\quad k=0,1,\ldots,2n-1\]
Now we just have to show that \(-z=z^*\):
\[\large\begin{align*}
-r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2\pi k}{2n}i\right)&=r^{\frac{1}{2n}}\exp\left(\frac{\theta_0+2n\pi+2\pi k}{2n}i\right)\\[2ex]
-\exp\left(\frac{\theta_0+2\pi k}{2n}i\right)&=\exp\left(\frac{\theta_0+(2n+2k)\pi}{2n}i\right)\end{align*}\]
Take the logarithm of both sides, using the principal branch, i.e. \(\ln \xi=\ln(se^{it})=\ln s+it\).
\[\large\begin{align*}
\ln(-1)+\frac{\theta_0+2\pi k}{2n}i&=\ln1+\frac{\theta_0+(2n+2k)\pi}{2n}i\\[2ex]
i\pi+\frac{\theta_0+2\pi k}{2n}i&=\frac{\theta_0+(2n+2k)\pi}{2n}i\end{align*}\]
and you can easily work out that the LHS=RHS.