Angular frequency is the way it is because $\sin$ and $\cos$ have simple derivatives in radians, and period $2\pi$ in radians. So, in order to convert an ordinary frequency into one $\sin$, $\cos$, and $\operatorname{e}^{ix}$ understand requires a multiplication by $2\pi$.
$\mathbf{k}$ is a vector and it's known as the wavenumber. It's components are related to wavelength in different directions in the same way period is related to angular frequency.
The space-time Fourier transform is just four Fourier transforms, one for each dimension. Traditionally the sign convention is chosen so that a wave with angular frequency $\omega$ propagates in the direction $\mathbf{k}$ points. That means that:
$$\begin{align} \tilde{f}(\omega,\mathbf{k}) &\propto \int f(t, \mathbf{x}) \operatorname{e}^{i\omega t - i \mathbf{k}\cdot\mathbf{x}} \operatorname{d} t \operatorname{d}^3x \\
f(t, \mathbf{x}) & \propto \int \tilde{f}(\omega,\mathbf{k}) \operatorname{e}^{-i\omega t + i\mathbf{k}\cdot \mathbf{x}} \operatorname{d}\omega \operatorname{d}^3k, \end{align}$$ where the constants of proportionality are chosen by convention, and different people use different conventions. I, personally, prefer the symmetric/unitary convention.
When you work in cylindrical and/or spherical coordinate systems you complicate matters a bit. Working in cylindrical coordinates moves you from straight Euclidean Fourier transforms into the realm of Hankel transforms in the radial direction and a discrete Fourier series in the aziumuthal angle. In the case of spherical coordinates, the radial transform becomes a spherical bessel function version of the Hankel transform and the angular coordinates become an expansion in spherical harmonics. These are all examples from linear algebra of writing a function in a particular orthonormal basis.
First, an example with simple exponential
We start off by a simplified example with the determining the Discrete-Time Fourier Transform (DTFT) of the exponential function given by
$$x_1[n]=e^{i\Omega_0n} \tag{1}.$$
Using the Fourier Transform
$$\mathcal F\{x_1[n] \} \tag{2}=X_1(\Omega)=\sum_{k=-\infty}^{\infty}x_1[k]e^{-i\Omega k}$$
we can calculate $\mathcal F \{ x_1[n]\}$ by substituting $x_1[n]$ and simplify the expression as follows:
$$\sum_{k=-\infty}^{\infty}e^{i\Omega_0 k}e^{-i\Omega k}=\sum_{k=-\infty}^{\infty}e^{i(\Omega_0-\Omega)k} \tag{3}$$
Then, by defining $\Omega'=\Omega - \Omega_0$ (observe the changed order) so that
$$\sum_{k=-\infty}^{\infty}1\cdot e^{-i\Omega'k} \tag{4}$$
we cleary see that it's now just the DTFT of a constant $1$, i.e. $\mathcal F \{1 \}$, which is given by
$$\mathcal F \{1 \}=2\pi \delta(\Omega') \tag{5}$$
and $2\pi \delta(\Omega-\Omega_0)$ in our case. Now, why is this? Where does the $2\pi$ come from?
So we have assumed that $\mathcal F \{1 \} = 2\pi\delta(\Omega)$, which defines the constant $1$ as a function $x_2[n]=1$. If we use the Inverse Discrete-Time Fourier Transform (IDTFT)
$$\mathcal F^{-1} \{ X_2(\Omega) \} = \frac{1}{2\pi} \int_{-\pi}^{\pi} X_2(\Omega)e^{i\Omega n} d\Omega \tag{6}$$
and substitute $X_2(\Omega)$ for our assumed $2\pi\delta(\Omega)$ we get
$$\frac{1}{2\pi} \int_{-\pi}^{\pi} 2\pi\delta(\Omega)e^{i\Omega n} d\Omega \tag{7}$$
which, since $\delta(\Omega)$ is "on" or $1$ only for $\Omega=0$, evaulates to
$$\frac{2\pi}{2\pi} e^{i\Omega 0} = 1. \tag{8}$$
We have thereby shown that the DTFT of a constant $1$ is equal to $2\pi\delta(\Omega)$.
If we now continue from equation $(4)$, we have that the resulting DTFT of an exponential $x_1[n]=e^{i\Omega_0n}$ becomes
$$\mathcal F\{x_1[n] \} = \sum_{k=-\infty}^{\infty}1\cdot e^{-i\Omega'k} = 2\pi\delta(\Omega') = 2\pi\delta(\Omega - \Omega_0) \tag{9}.$$
Secondly, show Discrete-Time Fourier Transform of a sine
We have the discrete-time function
$$y[n]=\sin(\Omega_0n+\phi) \tag{10}.$$
Observe that the $\phi$ causes a time-shift and can be handled separately. Therefore consider, for now, only the function
$$x[n]=\sin(\Omega_0n) \tag{11}.$$
Using Euler's formula we can write the sine as exponentials in the form
$$x[n]=\frac{1}{2i} \left ( e^{i\Omega_0n} - e^{-i\Omega_0n} \right ) \tag{12}$$
Using the DTFT of $x[n]$ we get
$$\mathcal F\{ x[n] \} = \sum_{k=-\infty}^{\infty}\frac{1}{2i} \left ( e^{i\Omega_0k} - e^{-i\Omega_0k} \right ) e^{-i\Omega k} \tag{13}$$
which can be rewritten as
$$\frac{1}{2i} \left [ \sum_{k=-\infty}^{\infty}e^{-i(\Omega-\Omega_0)k} - \sum_{k=-\infty}^{\infty}e^{-i(\Omega+\Omega_0)k} \right] \tag{14}.$$
We can now see that each term in equation $(14)$ is on the form in equation $(9)$ and therefore we get
$$\frac{1}{2i} \left [ 2\pi\delta(\Omega-\Omega_0) - 2\pi\delta(\Omega+\Omega_0 \right] \tag{15}$$
and
$$i\pi \left [ \delta(\Omega+\Omega_0) - \delta(\Omega-\Omega_0) \right] \tag{16}$$
if we shift the order of the terms to compensate the negative sign when $i$ jumps up from the denominator.
To consider the time-shift $\phi$ we mutltiply our results with $e^{i\Omega\phi}$ and finally get
$$e^{i\Omega\phi} \cdot i\pi \left [ \delta(\Omega+\Omega_0) - \delta(\Omega-\Omega_0) \right] \tag{17}$$
I'm not a hundred percent certain of my answer and I don't fully understand how to get the general form in the questions asked. If someone has edits to suggest, please do so.
The solutions are inspired and based on the following YouTube videos:
Best Answer
The fourier transform is linear. This implies that if you multiply by a constant in the time domain, you multiply by the same constant in the frequency domain.