Physical meaning of Fourier transform of an electric field? If I take the Fourier transform of the autocorrelation of a signal in time, I will get the power spectral density. 
But if I just have a signal of the electric field, $E(t)$, in time, then the intensity, $I(t)$, would be given by:
$$ I(t) = \frac{1}{2} c \epsilon_0 E(t)^* E(t) $$
Although what is the physical meaning of the power spectrums of $E(t)$ and $I(t)$, respectively, and their differences? I was told that if this electric field is caused by a particular dipole transition (e.g. methanol 6.7 GHz line) that the half-width at half-maximum, $\Delta \nu$, of the power spectral density would correspond to the frequency bandwidth of the photons due to the uncertainty principle (i.e. the photons themselves would have a frequency centred at 6.7 GHz with a bandwidth of $\Delta \nu$). Although why is this true? Wouldn't this power spectrum only indicate the frequency of the signal itself and not the photons comprising it?
 A: 
If I take the Fourier transform of the autocorrelation of a signal in time, I will get the power spectral density.

It so happens that the autocorrelation function is a Fourier transform pair of the power spectral density.  This is not to say that the only way to calculate the power spectral density is from the autocorrelation function.
As I stated at https://physics.stackexchange.com/a/309544/59023, the power spectral density, $S_{k}$, is proportional to the square of the magnitude of the Fourier transform of a signal, i.e., $S_{k} \propto \lvert X_{k} \rvert^{2}$.

Although what is the physical meaning of the power spectrums of $E(t)$ and $I(t)$, respectively, and their differences?

First, let me use the generic symbol $X_{k}$ to represent the Fourier transform of the time domain signal $x_{n}$.
The words power spectrum are somewhat ambiguous here.  In principle, one can compute a power spectrum (i.e., respective value vs. frequency) from each component of $\mathbf{E}$ or its magnitude.  One can also compute the amplitude spectra, $A_{k} \propto \lvert X_{k} \rvert$, of the signal.
In the following, I will assume you are asking about $S_{k}$ and not $A_{k}$ for each of these.
The power spectrum of $\mathbf{E}$, whether of components ($E_{j}$) or vector magnitude ($\lvert \mathbf{E} \rvert$), describe the power of the field as functions of frequency with units (if properly normalized) of (V m-1)2 Hz-1.  This is useful when trying to determine whether there exists, e.g., a wave at a given frequency which would show up as peak above the backgroun in $S_{k}$.  If the oscillations exist only along the x-component of $\mathbf{E}$ (i.e., a longitudinal, electrostatic oscillation), then the spectrum of both $\lvert \mathbf{E} \rvert$ and $E_{x}$ would show a frequency peak but not $E_{y}$ or $E_{z}$.
The intensity, as you have written it, is just the field energy density multiplied by a constant.  Thus, the power spectrum of $I$ would be qualitatively similar to that of $\lvert \mathbf{E} \rvert$.

Although why is this true? Wouldn't this power spectrum only indicate the frequency of the signal itself and not the photons comprising it?

I am not sure about what you were told and whether you are correctly conveying that information.  A discrete Fourier transform or DFT (i.e., what you use in practice on real signals through algorithms like the FFT) is not the same as a continuous Fourier transform (CFT).  In a DFT, the frequency bin width is defined as:
$$
\Delta f = \frac{ f_{s} }{ 2 \ N } \tag{1}
$$
where $f_{s}$ is the sample rate of the signal [e.g., vectors per second] and $N$ is the number of individual points used in the DFT.
In a CFT, the minimum $\Delta f$ is mathematically zero (i.e., infinitesimally small) but quantum shows us that energy/momentum are quantized and thus have discrete values.  Therefore, there are physical limits on the lower bound of $\Delta f$.  In this case, a variant of the uncertainty principle is applicable, called the time-energy uncertainty principle, which is roughly given as:
$$
\Delta E \ \Delta t \geq \frac{\hbar}{2} \tag{2}
$$
where $\hbar$ is the Planck constant and $\Delta Q$ is the minimum resolution of quantity $Q$.
Thus, the transition has a known energy change but we cannot know this better than that given by Equation 2.  For photons, we can directly convert energy to frequency with some constants, i.e., $E = h \ \nu$, thus we have the limitation on the frequency resolution of the emitted photons.
