How to show the field operator creates a particle at known position in Fock space

I'm very confused by something I saw in Susskind's Advanced Quantum Mechanics Lecture 6. He introduces Fock space $F$, defines the creation/annihilation operators $a^+_n,a^-_n$ on it (in terms of their action on the basis states $\lvert n_1n_2...\rangle$) and then defines the field operator on $F$

$$\Psi^+(x) = \sum_n \psi_n^*(x) a^+_n$$ where $\psi_n(x)$ is the nth energy eigenfunction for the single particle space $H$. He then attempts to show that $\Psi^+(x)\lvert 0 \rangle$ is the state with one particle at position $x$, that $\Psi^+(y)\Psi^+(x)\lvert 0 \rangle$ is the state with two particles at positions $x$ and $y$, and so on. Unfortunately I can't follow the reasoning.

First he shows that if $a^+,a^-$ are the ladder operators on the single particle space $H$, and $\Psi^+(x) = \sum_n \psi_n^*(x) (a^+)^n$ then $\Psi^+(x)\lvert 0 \rangle = \lvert x\rangle$, i.e. this operator on $H$ maps the lowest energy eigenstate to the state of know position $x$. I followed that bit. But then he goes on to say that since $\Psi^+(x), \Psi^+(y)$ commute that

$$\Psi^+(y)\Psi^+(x)\lvert 0 \rangle = \lvert x,y\rangle$$

But this makes no sense (to me) because the LHS is a state in $H$ and the RHS is a state in $H^2$. So my questions are:

1. Does this last equation actually mean something?


2. How do we achieve the real goal, which is to show that this equation is true in the fock space $F$?