Let me answer the $\omega$ vs $\xi$ question first. It's really a stylistic choice, although for physicists and engineers, the symbols do hold some meaning. $\omega$ is the angular frequency, $\xi$ here represents the frequency (because you can't use $f$, it's taken), and they are related via $\omega=2\pi\xi$. And no, it does not matter what you take when you do transformations.
Now let me answer the $2\pi$ factor part. I will try to be detailed, because this is a fascinating piece of mathematics!
Short answer : $1/2\pi$ is a scaling factor. You want the $2\pi$ either in the exponential itself (to scale it up), or you want it with the measure $\mathrm dx$ (to scale it down) which you can then pull out of the integral.
Long answer : Fourier transform is part of what we call harmonic analysis, and is a map that takes a good enough function $f$ on a (locally compact abelian) group $G$ and spits out a function $\hat f$ on $\hat G$ (which is the dual group of $G$).
Let me outline precisely what is being done (although i would be playing hard and loose with the conditions on functions) and you will see where the factor comes in.
Let $G$ be a locally compact abelian group (LCAG in short), by which I mean that $G$ is a an abelian group, and it has a topology on it (such that multiplication and inverse are continous with respect to the topology) which is locally compact. Also let $$\mathbb T=\{z\in\mathbb C\mid |z|=1\}$$ be the circle group. We then define $\hat G$ to be the set of all continuous homomorphisms from $G$ to $\mathbb T$, that is $$\hat G=\{\chi:G\to\mathbb T\mid \chi\text{ is continuous and a homomorphism}\}$$ $\hat G$ can be given a topology (called the compact-open topology) and it turns out that $\hat G$ is also a locally compact abelian group. This group is called the dual group of $G$ and the $\chi$ are called the characters of $G$. [As a side note, one of the greatest achievements of this field was proving that $G\cong \hat{\hat G}$, the so called Pontryagin Duality].
Now it turns out that LCAGs have a nice property : They admit a Haar measure $\mu$ on it! (This measure is unique upto scaling, and this scaling property is where the $2\pi$ factor will play a role). This allows us to define sizes of sets in the group and hence allow us to integrate stuff.
This is where we get into familiar territory. For a good enough function, we define the fourier transform of a function $f$ on $G$ to be the following
$$\hat f(\chi)=\int_Gf(x)\overline{\chi(x)}\mathrm d\mu(x)$$
Turns out, again for good enough functions $f$, we can define a fourier inverse
$$f(x)=\int_{\hat G}\hat f(\chi)\chi(x)\mathrm d\nu(\chi)$$ (Don't worry about the $\nu$ much, we will see in an example how it looks).
All is well and good, how does all this abstract machinery even look like for us? Let's start with an easy example to get familiar with how this relates to what we know as classical fourier series/transform.
Example 1 : Let $G=\mathbb T$. What is $\hat G$? Well, we need to find maps $\chi:\mathbb T\to\mathbb T$ that are continuous homomorphisms. Note that all maps $z\mapsto z^n$ (for $n\in\mathbb Z$) satisfy this property. A little more work shows that these are all, there is nothing else. Label these maps as $\chi_n$. Note that if you know $\chi_1$, you know all the other maps (why?). Thus is is an infinite cyclic group with one generator, and hence is $\mathbb Z$. Thus $\chi_n(x)=e^{2\pi inx}, x\in [0,1)$ where I have considered $\mathbb T$ as $\mathbb R/\mathbb Z$ to write $x$ as a real number (note the choice I made of the $2\pi$ factor in the exponential. If I had not included that, the interval for $x$ would have become $[0,2\pi)$ instead [note the scaling]).
Concluding the example, we now have that for a good enough function $f$ on the circle, we can write its fourier transform $$\hat f(n):=\hat f(\chi_n)=\sum_{j\in\mathbb Z}f(n)e^{-2\pi inx}$$ which is just our good old fourier series!
Example 2 : Now let $G=\mathbb R$. What is $\hat{\mathbb R}$? This is a fun result : $\hat{\mathbb R}\cong \mathbb R$! The fact is proven here, and the characters are $\chi_y(x)=e^{2\pi ixy}$ where $y\in\mathbb R$ (again note the $2\pi$ factor, it could very well not have been there, we will revisit this very soon).
For now, take the measure on $\mathbb R$ as the ordinary Lebesgue measure. Then our fourier transform looks like $$\hat f(y):=\hat f(\chi_y)=\int_{\mathbb R}f(x)e^{-2\pi ixy}\mathrm dx$$ which is what you mentioned in your second equation. How about the inverse transform? Well, let's put $f(x)=e^{-\pi x^2}$. We then have $$\hat f(y)=\int_{\mathbb R}e^{-2\pi ixy-\pi x^2}\mathrm dx$$ Completing the square in the exponential gives $\hat f(y)=e^{-\pi y^2}=f(y)$ and since $f$ is even, we have $$f(x)=\int_{\mathbb R}e^{2\pi ixy-\pi y^2}\mathrm dy$$ and thus the measure on the dual group has no scale factor (that is, the measure on $\mathbb R$ and the measure on $\hat{\mathbb R}$ are the same, the Lebesgue measure itself).
Now note that there were two choices we made in the above. One was choosing how we write the characters, and the other was the measure on $\mathbb R$. How about we change these and see what happens.
Suppose we change the way we write characters. Say we do $\chi_y(x)=e^{ixy}$. What changes? Well, keeping the measure fixed, we have $$\hat f(y)=\int_{\mathbb R}f(x)e^{-ixy}\mathrm dx$$ Now again going through plugging in $f(x)=e^{-\pi x^2}$, we have $$\hat f(y)=\int_{\mathbb R}e^{-ixy-\pi x^2}\mathrm dx$$ which gives $$\hat f(y)=e^{\frac{-y^2}{4\pi}}=f(x/\sqrt{2\pi})$$ The glimpses of that extra $2\pi$ are almost here. Plugging this in the fourier inversion tells us that $$\int_{\mathbb R}e^{ixy}\hat f(y)\mathrm dy=\int_{\mathbb R}e^{ixy-\frac{y^2}{4\pi}}\mathrm dy=2\pi e^{-\pi x^2}=2\pi f(x)$$
which is exactly the factor of $2\pi$ extra! To remedy this, all we have to do this is consider the measure on $\hat{\mathbb R}$ to be $\mathrm dy/2\pi$ which essentially modifies the fourier inversion formula to $$f(x)=\int_{\mathbb R}\hat f(y)e^{ixy}\frac{\mathrm dy}{2\pi}=\frac{1}{2\pi}\int_{\mathbb R}\hat f(y)e^{ixy}\mathrm dy$$
Well, this is close to what you mentioned in your equations, but not quite. This is the fourier inversion, not the fourier transform. Well, that's an easy fix because we saw how things behave. You choose your characters like $\chi_y(x)=e^{iyx}$ and choose your measure on $\mathbb R$ as $\frac{\mathrm dx}{2\pi}$ and the fourier transform and inversion becomes $$\hat f(y)=\frac{1}{2\pi}\int_{\mathbb R}f(x)e^{-ixy}\mathrm dx$$ $$f(x)=\int_{\mathbb R}\hat f(y)e^{ixy}\mathrm dy$$
Exercise : Show that if you choose your characters as $\chi_y(x)=e^{iyx}$ and the measure on $\mathbb R$ as $\frac{\mathrm dx}{\sqrt{2\pi}}$, then your fourier transform and inversion measures look the same.
