downsampling an fft signal
Downsampling and spectral stretching is earlier to show in 1D. I'll also analyze with the Fourier transform and discrete-time Fourier transform, not a DFT. The results are the same except for aliasing in time. 

To downsample by an integral multiple is to increase the sample period of a continuous-time signal $x(t)$ by an integer $M : T → M T$ . The replication period in the frequency domain is reduced by the same multiple. Rather than the first replica being centered at $\omega = 2\pi/T$ , the first replica of the down-sampled spectrum is centered at $\omega = 2\pi/MT$. Moreover, the amplitude is scaled by $1/M$. The figure illustrates $M = 2$. Sequencing the original series converts $x(nT ) → x[n]$. To downsample the sequence is to introduce a stride $M$ when reading off the values: 

$$
x^{\downarrow} [n] = x[nM ]
$$

The effect on the angular sample frequency can be shown by a units analysis,

$$
\left(\frac{φ^↓\; \text{radians}}{1\; \text{sample}}\right)
=
\left(\frac{ω\; \text{radians}}{1\; \text{sec}}\right)
\left(\frac{T\; \text{sec}}{1\; \text{step}}\right)
\left(\frac{M\; \text{step}}{1\; \text{sample}}\right)
$$

or $φ^{\downarrow} = \omega M T$ . The range $ω ∈ [0, 2π/M T ]$ corresponds to $φ^↓ ∈ [0, 2π]$, thus $φ^↓$ is a stretched version of $ω$. The figure below compares $x[n]$ to $x^↓ [n]$ and $φ$ to $φ^↓$ . $x^↓ [n]$ has fewer samples than $x[n]$, and has a broader spectrum when plotted in $φ$ because the radians per sample has increased by $M$.



So in the 1D analogy to your case, refer to the continuous-frequency spectra $\omega$ above. Downsampling by 2:1 creates a spectral replica centered at $2\pi/MT$, $M=2$. That is what your first lines of code do (in 2D). Then, the spectrum is stretched so that the periodicity of the spectrum is restored to $2\pi$. This is shown to the right.