OK there is a lot here to unpack.
First off, what do we mean by a particle in a QFT?
Well the formal definition is something like, "particle states fall into representations of the Poincaire group."
In perhaps more physical terms, a particle, in QFT, is defined by a momentum eigenstate (which might also have associated spin). You actually run into seriously trouble if you try to take the position eigenstates of a particle too seriously [briefly: if you try to localize a particle to a very small region within its own Compton wavelength, the uncertainty principle allows for particle creation, and so you get a large number of particle and antiparticle pairs being created if you try to localize the particle and the calculation blows up in your face], and so we never really think about the position of a relativistic quantum particle.
To define a particle we usually look at it when its momentum is zero, in which case a particle is defined by its mass (energy at 0 momentum) and its spin (angular momentum at zero momentum). String theory recreates particles in this sense--there are states in string theory that have mass and spin, that therefore will look like particles to a low energy observer. Furthermore, we can compute scattering of strings, which will look like scattering of particles to low energy observers who can't see that the string is stringy.
A theory of a free string (ie a worldsheet with a trivial topology) does not reproduce the full multiparticle Fock space of a quantum field theory. Thus, we can't construct a full quantum field operator $\phi(x)$ from a single free string. Instead, a theory of a single free string reproduces the single particle states of an infinite number of quantum field theories.
In other words, when you quantize the string (on a worldsheet with trivial topology), you end up with a theory of a single string. A single string looks like a single particle to a low energy observer (one who can't probe the 'stringiness' of the string). The key is that there are many different single particle states that a string might be in, characterized by mass and spin. The string has some internal structure--the low energy observer can't directly probe this structure, and so only sees a point particle--but the internal structure shows up as different masses or spins.
Let's say it one more time for good measure. A quantum field theory is a theory of an indefinite number of identical particles. When you quantize a string, you get a theory of a single string that can look like any one of an infinite tower of particles, with different masses and spins.
OK in math here is what I mean:
You can think of the string as a quantum theory, with a position operator $X^\mu$ and a momentum operator $p^\mu$. Actually its more convenient to work with the fourier transform of the position operator, we call the coefficeints $\alpha$. The Hamiltonian is given by [this is for the open string in light cone gauge, being lazy about some factors of $\alpha'$ and $2\pi$]
\begin{equation}
H = \frac{1}{2p^+} \left( \sum_{i=1}^{D-2}\frac{1}{2} p^i p^i + \sum_{n\neq 0} \alpha_n^i \alpha_{-n}^i\right)
\end{equation}
The first term, summing over $p^i p^i$, is essentially just the hamiltonian of a free particle, and is therefore not super interesting.
Perhaps you will notice that the second term, with the $\alpha_n$, is very similar to a sum of harmonic oscillator hamiltonians written in terms of the creation/annhilation operators. In fact that relationship can be made very precise, look at those notes I linked to for details.
The spectrum of the hamiltonian is thus made of a vacuum state with a bunch of equally spaced harmonic oscillator excited states living on top of it. Actually this isn't quite correct--there is actually an infinite set of vacuum states labeled by $p^\mu$
\begin{equation}
|0;p^\mu \rangle
\end{equation}
These are the eigenstates of the first term in the hamiltonian. The label $p^\mu$ is just the "center of mass motion" of the string. We can safely ignore it for the purposes of this question, and similarly we can ignore the first term in the hamiltonian above.
The interesting thing are the excited states. Just like a harmonic oscillator, you get the excited states by acting with the creation and annhilation operators. Well, we could construct proper creation and annhilation operators, but in this context it is more convenient to use the level operators, which differ from the by a normalization factor and its important to keep track of the $n$ label, however the level operators are morally very similar to creation/annhilation operators ($n<0$ correspond to creation operators). Thus the excited states are things like
\begin{equation}
\alpha_{-1}^i |0\rangle, \ \ \alpha_{-2}^i|0\rangle, \ \ \alpha_{-1}^i \alpha_{-1}^i |0\rangle, \ \ \alpha_{-1}^i \alpha_{-10}^i \alpha_{-32}^i |0\rangle
\end{equation}
Now this looks an awful lot like a Fock space from quantum field theory, where the creation/annhilation operators are creating/annhilating particles. So it looks like, by analogy, when I act with two level operators I will create two particles.
BUT THAT IS NOT WHAT IS GOING ON!!!!!!!
We only have one string. We are not creating particles, we are not creating strings. We are creating excited states of a single string.
How do we determine what this state looks like to a low energy observer? Well we compute the mass (ie the energy, assuming the center of mass motion is zero), and we compute the spin (ie the angular momentum, assuming the orbital angular momentum aka center of mass motion is zero).
How do we compute the energy and the angular momentum?
Well actually we have the position $X^\mu$ and momentum $p^\mu$ operators in the theory already, so we just use the normal definitions: $p^0$ is the energy of the string and you can use $X$ and $p$ to construct an angular momentum operator. Acting on the excited states above, you find that the states differ by mass and spin.
That's the outline, obviously I am skimming over many details, but honestly going through all of it slowly would take at least a full lecture in a string theory course, so I would recommend reading Tong's notes (link below) if you want more detail.
Anyway, to very very briefly address your other ponits:
How do we get QFT out? Well in this language it looks like honest way to do it would be to have a theory of an indefinite number of identical strings. This is known as string field theory, and is very hard. However, the modern perspective is that the strings are something of a red herring and are not the real fundamental degrees of freedom in the theory, so maybe string field theory is not the way to go in general.
Wait, but we can get QFT out, right? OK, yes it is true we can reproduce results of QFT from string theory, otherwise what would be the point? We can compute scattering of strings (it turns out what looks like two strings scattering can really be thought of as one string worldsheet with a tear in it). So we get an S matrix for strings, we can take a low energy limit of that and get an S matrix for particles. We can then write down a QFT lagrangian that reproduces the same low energy S matrix. By using methods along these lines we find that the low energy results of string theory are reproduced by writing down supergravity actions.
What does $X^\mu (\sigma,\tau)|0 \rangle$ mean? First I disagree with you that $\sigma$ and $\tau$ are "completely fake things", they are coordinates on the worldsheet. The worldsheet of the string is real (or at least, it is real if the string is real :)), so $\sigma,\tau$ are labelling real things. Also I disagree with you calling $\phi(x) |0\rangle$ a position wave function in QFT. I would call it an insertion of the operator $\phi$ at the point $x$. You are probing the quantum field at the point $x$. Similar, by acting with the operator $X^\mu(\sigma,\tau)$, you are probing the string by poking it at a point on the strings worldsheet labeled by $\sigma,\tau$.
Reference:
A very good set of lecture notes by David Tong can be found here.
See chapter 2 especially (he is doing the closed string but it's morally similar for what you are interested in).
It depends, roughly speaking, on what is the operator that you interpret as the number operator of the theory. It is customary to choose $N$ as the operator such that: a vector $\phi$ of the Fock space is in the $n$-particle sector if $N\phi=n\phi$, i.e. if it is an eigenvector of the number operator $N$ with eigenvalue $n\in\mathbb{N}$.
A possible choice is $N=\int a^\dagger a$. Observe that the number operator is not just $a^\dagger(\cdot)a(\cdot)$, but the integration of this over the variable. I am not specifying the variable because you can think of $a^{\#}(\cdot)$ as the creation/annihilation operators in the position or momentum representation, it does not matter (the two representations of $N$ are unitarily equivalent, and related by the Fourier transform).
This choice is related to the mathematical theory of Fock spaces, and depends, roughly speaking, from the fact that $a^\#(p)$ (or $a^\#(x)$) are not operators, but make sense as an operator only when they are integrated with a square integrable function, e.g.
$$a^{\dagger}(f)=\int_{-\infty}^\infty f(p) a^\dagger(p)dp$$
with $f(p)\in L^2$ is an operator on the Fock space ($a^\#(p)$ are often called operator valued distributions). You superposition for example is not a vector of the Fock space, because $e^{-px^2}$ is not square integrable w.r.t. $p$.
You see that the choice above implies: $N\lvert 0\rangle=0$, and formally (it is easy to see using the canonical commutation relations):
$$N\int_{-\infty}^\infty dp e^{-px^2}a_p^\dagger\lvert 0\rangle=\int_{-\infty}^\infty dp e^{-px^2}a_p^\dagger\lvert 0\rangle$$
So your "vector" (as I said it does not belong to the Fock space) has only one particle (and the vacuum zero). It is also easy to see (again using the commutation relations) that for any $n\in\mathbb{N}$, $f\in L^2(\mathbb{R}^{n})$ the vector:
$$\phi_n=\int f(p_1,\dotsc,p_n)a^\dagger(p_1)\dotsm a^\dagger(p_n)dp_1\dotsm dp_n\lvert 0\rangle$$
belongs to the $n$-particle subspace, i.e. $N\phi_n=n\phi_n$.
Best Answer
It works because collider experiments do not measure (x,y,z,t). They measure (p_x,p_y,p_z,E). The calculations are done for point particles entering Feynman diagrams but the numbers that predict measurements are not dependent on space time, but on energy momentum.
No experiment can measure the localization of an individual interaction with the accuracy necessary to see effects of spatial uncertainty: the incoming protons have the Heisenberg uncertainty even if they were measured individually and not as a beam, and the same would be true for the outgoing particles that would have to be extrapolated back to the vertex. Any predictions on the localization of the interaction in the beam crossing region would fall within these combined HUP uncertainties, imo of course.