One way to look at quantum field theory is that it is the theory of an indefinite number of identical particles.
Before explaining that, let's recall that there are two common bases we can think about in quantum field theory. We will assume we are working with a free theory of a real scalar field, for simplicity.
First, we can work in terms of field eigenstates $|\phi(x)\rangle$, where the field operator $\hat{\phi}(x)$ takes on a definite value
$$
\hat{\phi}(x) |\phi(x) = \phi(x) |\phi(x)\rangle
$$
The wave functional you've defined is expressed in this basis
$$
\Psi[\phi(x,t)] = \langle \phi(x,t) | \Psi \rangle
$$
Second, we can work in terms of a Fock space basis, or space of states with a definite number of particles. For a scalar field, the basis states are:
- Vacuum: $| 0 \rangle$
- Single particle states: $|p\rangle$
- Two particle states: $|p_1\rangle | p_2 \rangle = |p_1, p_2\rangle$
- etc
In Fock space, it is easy to define the action of creation and annihilation operators $\hat{a}_p^\dagger, \hat{a}_p$, so for example,
$$
\hat{a}_{p_1}^\dagger \hat{a}_{p_2}^\dagger \cdots \hat{a}_{p_n}^\dagger | 0 \rangle = |p_1 p_2 \cdots p_n \rangle
$$
We can also define a number operator $\hat{N}$ to measure the number of particles in each state
$$
\hat{N} = \int \frac{d^3 p}{(2\pi)^3} \hat{a}_p^\dagger \hat{a}_p
$$
where
$$
\hat{N} |p_1 p_2 \cdots p_n \rangle = n |p_1 p_2 \cdots p_n \rangle
$$
In this framework, the statement that QFT describes an indefinite number of identical particles is the most clear; a general state will not be an eigenstate of $\hat{N}$, and will involve a superposition of states from different sectors of Fock space, with different numbers of particles.
Now, we can express the field operator in terms of the creation and annihilation operators
$$
\hat{\phi}(x) = \int\frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left[\hat{a}_p e^{i p x} + \hat{a}_p^\dagger e^{- i px}\right]
$$
Using this expression, we can derive the fact that the field and number operator do not commute
$$
[\hat{\phi}(x), \hat{N}] \neq 0
$$
Therefore, we cannot simultaneously diagonalize the field and number operators. In other words, if we work on a field basis, then there is not a definite particle number. Or, in general, we should expect a field state $|\phi(x)\rangle$ to be a superposition over all sectors of Fock space
$$
|\phi(x) \rangle = c_0 |0\rangle + \int \frac{d^3 p}{(2\pi)^3} c_p |p\rangle + \int \frac{d^3 {p_1}}{(2\pi)^3} \int \frac{d^3 {p_2}}{(2\pi)^3} c_{p_1, p_2} |p_1, p_2\rangle + \cdots
$$
where the $c_k$ are c-numbers representing coefficients in a superposition of states. (You can work these out explicitly for a free theory, there's an expression on wikipedia, albeit in different notation).
The upshot of the above is that it's going to be difficult to use the wave functional (as a function of $\phi(x,t)$) to talk about states with definite numbers of particles.
Now, let's go back to the double-slit experiment.
Since we are dealing with a bosonic field theory, there are multiple versions of this experiment we can describe.
The classical version of the double slit experiment is to send a classical wave through the slits and see an interference pattern. This can be described easily using field eigenstates $|\phi(x)\rangle$. We simply start with a classical plane wave state and evolve it, and we will see normal, classical interference effects. One way to view what's going on quantum mechanically is that because we are working with a bosonic field, we can put many particles into the same state. Many particles acting in a similar way will behave like a classical field. The formal version of that is what we saw above, that a field eigenstate will involve a superposition over states with arbitrarily large numbers of particles. This is a boring case.
The quantum version of the double slit experiment is to send a single particle through the slits at one time. We then repeat this experiment many times and build up the superposition.
Now, since we're working with a free theory, you expect "many particles sent one after another" to behave the same way as "many particles sent at once" since the particles don't interact with each other. So from our fancy field theory perspective, we can see how the quantum version will have to give the same result as the classical version if we use many particles.
But if you want to directly see what's going on in the quantum case, one particle at a time, then what all of this is saying is that the wave functional is the wrong tool to understand this physical situation. The wave function makes it easy to understand field eigenstates or states that are approximately field eigenstates. But field eigenstates are going to involve a superposition including states with large numbers of particles, which are precisely what we don't want to deal with to understand the quantum double-slit experiment. The easier way to approach this problem is to restrict ourselves to the single-particle subspace of the full Fock space and ask what happens there. Then, we essentially reduce ourselves to the case of single-particle quantum mechanics (see, eg, Tong's QFT notes, Section 2.8.1).
If you are really determined, you can describe single particle states using a wave function, at least in a free theory. (Here, I'm following wikipedia). You start with the ground state wave functional
$$
\Psi_0[\phi] = \langle 0 | \Psi \rangle = {\rm det}^{1/4} \left(\frac{K}{\pi}\right) \exp\left(-\frac{1}{2} \int d^3 x \int d^3 y \phi(x) K(x,y) \phi(y) \right)
$$
where the covariance $K$ is
$$
K(x,y) = \int\frac{d^3 k}{(2\pi)^3} \omega_k e^{i k(x-y)}
$$
Then, you can define a wave functional for a one-particle state as
$$
\Psi_1[\phi] = \langle p | \Psi\rangle = \left(\frac{2\omega_{p}}{(2\pi)^3}\right)^{1/2} \int d^3 y e^{- i p y} \phi(y) \Psi_0[\phi]
$$
From here, what you would do is (1) construct an appropriate superposition $\Phi$ of one particle states to mimic the set up of the double slit experiment, (2) compute what that superposition is explicitly in terms of $\Psi_1$, (3) compute how to express the operator corresponding to the particle's detected position on the screen $\hat{X}$ in terms of the field operator (to make things fun note there is no exact notion of a position operator in relativistic quantum field theory), (4) compute the expectation value $\langle \Phi | \hat{X} | \Phi \rangle$ as a functional integral
$$
\langle \hat{X} \rangle = \int D \phi_1 \int D\phi_2 \Phi[\phi_1]^\star \hat{X} \Phi[\phi_2]
$$
This is the mess you would need to go through to use the Schrodinger function to understand the double-slit experiment, which is why I would advocate not to use this tool for this problem. Somehow, amazingly, it must give the same answer as you would get using one-particle Fock space states -- I imagine the calculation involves many, many integrals collapsing into delta functions that cancel other integrals -- but I think the only point of doing this calculation would be a macho desire to prove it could be done, rather than deriving any insight about the physics. Unfortunately, this is a common situation in quantum field theory -- there are many ways that in principle should give the same result, but in practice only some methods are useful, and often you need to use different methods for different problems.
