Why zero padding the 2-d DFT interpolates images in spatial domain?
For the sake of simplicity, I'll explain the 1-D case; the 2-D case is completely analogous. Let $x[n]$ be a finite length sequence with $n\in[0,N-1]$. Its discrete Fourier transform (DFT) is

$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j\frac{2\pi}{N} kn}\tag{1}$$

The sequence $x[n]$ can be obtained from $X[k]$ via the inverse DFT (IDFT):

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn}\tag{2}$$

Note that the sequences defined by $(1)$ and $(2)$ are periodic with period $N$. Let's use $\tilde{X}[k]$ to denote the periodic sequence of DFT coefficients. For odd $N$, $\tilde{X}[k]$ just equals (the periodic continuation of) $X[k]$. For even $N$, we split the coefficient $X[N/2]$ in two equal terms $\frac12 X[N/2]$, such that for even $N$, the periodic sequence $\tilde{X}[k]$ is defined as

$$\tilde{X}[k]=\\\big[\ldots ,\underbrace{X[0],X[1],\ldots,\frac12 X[N/2],\frac12 X[N/2],\ldots,X[N-1]}_{N+1\textrm{ elements}},\ldots\big]\tag{3}$$

With that definition, the period of $\tilde{X}[k]$ is always odd. Let's denote that odd period by $2K+1$. For odd $N$ we have $K=(N-1)/2$, and for even $N$ we get $K=N/2$.

Using $\tilde{X}[k]$ and $K$ as defined above, we can rewrite $(2)$ as

$$x[n]=\frac{1}{N}\sum_{k=-K}^{K}\tilde{X}[k]e^{j\frac{2\pi}{N}kn}\tag{4}$$

With these preparations we can now interpret the coefficients $\tilde{X}[k]$ as Fourier coefficients of a continuous periodic function $x_c(t)$ with period $NT_s$, where $T_s$ is an arbitrary sampling interval:

$$x_c(t)=\frac{1}{N}\sum_{k=-K}^{K}\tilde{X}[k]e^{j\frac{2\pi}{NT_s}kt}\tag{5}$$

The sequence $x[n]$ can be obtained by sampling $x_c(t)$:

$$x[n]=x_c(nT_s)\tag{6}$$

But we can also obtain other, more densely sampled sequences by sampling $x_c(t)$. We choose the new sampling interval such that the new sequence is also periodic with period $M&gt;N$:

$$\hat{x}[m]=x_c\left(m\frac{NT_s}{M}\right)=\frac{1}{N}\sum_{k=-K}^{K}\tilde{X}[k]e^{j\frac{2\pi}{M}km}\tag{7}$$

The new length $M$ sequence $\hat{x}[m]$ can be interpreted as an interpolated version of the original length $N$ sequence $x[n]$.

Note that $(7)$ can be written as a length $M$ IDFT with a zero-padded sequence $\tilde{X}_0[k]$:

$$\begin{align}\hat{x}[m]&amp;=\frac{M}{N}\left(\frac{1}{M}\sum_{k=0}^{M-1}\tilde{X}_0[k]e^{j\frac{2\pi}{M}km}\right)\\&amp;=\frac{M}{N}\cdot \textrm{IDFT}_M\big\{\tilde{X}_0[k]\big\}[m]\tag{8}\end{align}$$

with

$$\tilde{X}_0[k]=\big[\tilde{X}[0],\ldots,\tilde{X}[K],\underbrace{0,\dots,0}_{M-2K-1\textrm{ zeros}},\tilde{X}[K+1],\ldots,\tilde{X}[2K]\big]\tag{9}$$

In terms of the original DFT coefficients $X[k]$, the zero-padded sequence $\tilde{X}_0[k]$ is given by

$$\tilde{X}_0[k]=\big[X[0],\ldots,X[(N-1)/2],\underbrace{0,\dots,0}_{M-N\textrm{ zeros}},\\\qquad X[(N+1)/2],\ldots,X[N-1]\big]\tag{10}$$

for odd $N$, and by

$$\tilde{X}_0[k]=\big[X[0],\ldots,\frac12 X[N/2],\underbrace{0,\dots,0}_{M-N-1\textrm{ zeros}},\\\qquad \frac12 X[N/2],\ldots,X[N-1]\big]\tag{11}$$

for even $N$.