What is the mathematical relationship between the wave functions of QM and the fields in QFT? To be more specific, say I have the wavefunction for the position of some arbitrary particle, but then I also have the QFT operator (is that the correct term?) modeling the position of that same particle.  What's the relationship between those two mathematical objects?  Like, could a 1-to-1 correspondence be reasonably created between the wave functions of QM and the operators of QFT?
Or, to put it in more conceptual terms, is it accurate to say that particles are excitations in quantum fields and the PDF's we get from squaring the wavefunctions in QM correspond to the physical distributions of the excitations in the fields?  The image I have in my head is of an object being dropped onto a clear sheet and creating ripples, and I imagine the shape of the ripples would correspond to the shape of the probability distribution for the particle, where the particle is just the ripples themselves.  I'm guessing the actual relationship is more complicated than that though.  Is it?  If so, in what way?
Edit: Looking at the answer to this question, as was suggested: Why we use fields instead of wave functions?
I get the impression that a quantum field is just the superposition of all the relevant wave functions, like, that the electron field is the superposition of the wavefunctions of all electrons.  Is that right?
 A: The problem connecting QFT and $N$-body quantum mechanics isn't so much with QFT but rather with relativistic field theories. For a non-relativistic theory the connection is actually quite straight forward, if we define the one-particle momentum states by
$$a^{\dagger}_\mathbf{p}|0\rangle=|\mathbf{p}\rangle$$ then the position basis from regular QM is simply $$|\mathbf{x}\rangle=\frac{1}{(2\pi)^{3/2}}\int d^3p \ e^{-i\mathbf{p}\cdot\mathbf{x}}a^{\dagger}_\mathbf{p}|0\rangle.$$ These are delta-normalized because of the commutation relations for $a_\mathbf{p}$ and so the regular QM wavefunction for a one particle state is $$\psi(\mathbf{x})=\langle\mathbf{x}|\psi\rangle=\frac{1}{(2\pi)^{3/2}}\int d^3p \ e^{i\mathbf{p}\cdot\mathbf{x}}\langle0|a_\mathbf{p}|\psi\rangle.$$ The problem comes with relativistic field theories where the measure $d^3p$ has to be replaced with $d^3p/2E_p$ for lorentz invariance and this ruins the orthogonality of the analogous position basis states, more precisely
$$\frac{1}{(2\pi)^3}\int \frac{d^3p}{2E_p}\ e^{i(\mathbf{x}-\mathbf{x'})\cdot\mathbf{p}}\neq\delta^3(\mathbf{x-x'}).$$ The way to think about this is that relativistic particles/fields have an inherent amount of spatial spread and so it's impossible to construct a completely localized state. Duncan briefly discusses this in section 6.5 of his textbook The Conceptual Framework of Quantum Field Theory in which he showed that you can't construct a one-particle state that is delta-localized for both the number density operator and the enery density operator.
A: When we do quantum field theory, we still are doing quantum mechanics. We don't  normally work a lot with a wave function explicitly because the wavefunction (other than perhaps the vacuum state) is quite complicated (and hard to prepare in many cases).
Nevertheless, all the postulates of Quantum Mechanics still hold, we just usually do not write out a single N-particle Hamiltonian, since we usually want to consider the case where the number of particles is changing.

Another way to understand why we have to introduce all this fancy new formalism is to realize that in "ordinary" N-particle quantum mechanics space and time are treated completely differently. In N-particle QM, Spatial position is an operator, but time is a real number.
If you try to promote time to an operator, you run into trouble. So, instead  we demote position to a number. But, we still need operators, and those are the fields; the fields are operator-valued functions of space and time (where both space and time are just real numbers).


What is the mathematical relationship between the wave functions of QM and the fields in QFT?

The fields are operators, so they act on wavefunctions.
For example, when we write a Klein-Gordan field operator like:
$$
\hat\phi(\vec r, t)\;,
$$
we mean that there is an operator at every point in space-time.
This operator can act on, say, the ground state wave function of the universe $|\Phi_0\rangle$ and create a new wavefunction:
$$
|\Psi\rangle(\vec r, t) = \hat\phi(\vec r, t)|\Phi_0\rangle\;.
$$


I get the impression that a quantum field is just the superposition of all the relevant wave functions, like, that the electron field is the superposition of the wavefunctions of all electrons. Is that right?

This is not the right way to say this.
It is perhaps helpful to think of particles as localized excitations associated with the field. This is nice, because it help us feel good about why, say, every electron is identical. We can say they are all identical because they are all excitations created by the same field operator $\hat \psi(\vec r,t)$, which itself obeys the Dirac equation.

Update to address one comment:
How do we recover the Schrodinger equation? We don't need to recover it, per se, we just need to write it in terms of the field operators. It looks the same:
$$
i\partial_t |\Psi\rangle(t) = \hat H |\Psi\rangle(t)\;,
$$
but here the Hamiltonian is written as an integral over all space of the operators. For example, for a free Klein-Gordan field:
$$
\hat{H} = \int d^3r \frac{1}{2}\left(
\dot{\phi(\vec r)}^2 + \nabla\phi(\vec r)^2+ m^2\phi(\vec r)^2\;,
\right)
$$
where the operators have no time dependence above since we are working in the Schrodinger picture now. (And I have dropped the "hats" from the operators.)
This Hamiltonian can be rewritten as:
$$
H = \int \frac{d^3p}{(2\pi)^3}\omega_p a^\dagger_{\vec p}a_{\vec p} + C\;,
$$
where $C$ is a numerical (non-operator) constant. And where:
$$
\phi(\vec r) = \int \frac{d^3p}{\sqrt{2\omega_{\vec p}}(2\pi)^3}\left(
a_{\vec p}e^{i\vec p\cdot \vec x}+ a_{\vec p}^\dagger e^{-i\vec p\cdot \vec x}
\right)
$$
