So let's start a step back because your coherent states are not normalized as I would normalize them.
Coherent states
The coherent states come from their response to the bosonic annihilator, $$\hat b |x , y\rangle = (x + i y) |x,y\rangle.$$From this one can derive that any particular one's representation among the number states must satisfy, $$\hat b~\sum_n c_n |n\rangle = \sum_n c_n \sqrt{n} |n-1\rangle=(x + i y) \sum_n c_n |n\rangle,$$giving the recursive relation that $c_n = \frac{x+iy}{\sqrt n}~c_{n-1}.$ Starting from $c_0$ we then find indeed the relation that $$|x,y\rangle = c_0~\sum_n \frac{(x + i y)^n}{\sqrt{n!}} |n\rangle.$$The remaining $c_0$ with the proper normalization gives $$\langle x,y|x,y\rangle = 1 = |c_0|^2 \sum_n \frac{(x-iy)^n(x+iy)^n}{n!} = |c_0|^2 \exp\big(x^2 + y^2\big).$$Choosing these to all have the same complex phase for their vaccum component finally yields,$$|x, y\rangle = \exp\left(-\frac12(x^2 + y^2)\right)\sum_n \frac{(x + i y)^n}{\sqrt{n!}}~|n\rangle.$$
So the question is, why does your expression have a leading $\pi^{-1/2}$ in it? That's because they resolve the identity in a somewhat weird way. What does that mean?
Resolving the identity
Suppose you have an expression for some average $\langle A \rangle.$ QM is very clear that this expression may be written based on its quantum state $|\psi\rangle$ as $\langle \psi|\hat A|\psi\rangle.$
But using the fact that $1 = \sum_n |n\rangle\langle n|,$ for example, we can insert these sums ad-hoc into that expression to find that in fact this expectation value also reads, $$\langle A \rangle = \sum_{mn} \langle\psi|m\rangle\langle m|\hat A|n\rangle\langle n|\psi\rangle = \sum_{mn} \psi^*_m~A_{mn}~\psi_n.$$ So that is the value of resolving the identity; it means that you can define this matrix $A_{mn}$ which fully specifies the action of $\hat A$ on the Hilbert space, recovering every single expectation value from the matrix.
Well we see something very similar when we look at the operator, $$\hat Q = \int_{-\infty}^\infty dx~\int_{-\infty}^\infty dy~|x,y\rangle\langle x, y| = \sum_{mn} \iint dx~dy~e^{-x^2-y^2}\frac{(x-iy)^m(x+iy)^n}{\sqrt{m!n!}} |m\rangle\langle n|.$$
At this point it is useful to shift to polar coordinates where $x + i y = r e^{i\theta},$ yielding $$\hat Q = \sum_{mn}\int_{0}^\infty dr~\int_0^{2\pi} r~d\theta~e^{-r^2}~\frac{r^{m+n} e^{i(n-m)\theta}}{\sqrt{m!n!}} |m\rangle\langle n|.$$ Note that the angle over $\theta$ integrates a sinusoid over one or more full periods and therefore vanishes if $m\ne n$; it is $2\pi$ if $m = n$, so we
must get:$$\hat Q = \pi\sum_{n}\int_{0}^\infty dr~2r~e^{-r^2}~\frac{r^{2n} }{n!} |n\rangle\langle n|.$$Substituting $u=r^2, du=2r~dr$ we find that this is:$$\hat Q = \pi\sum_{n}\frac1{n!}~|n\rangle\langle n|~\int_{0}^\infty du~e^{-u}~u^n.$$If you've never seen the gamma function before, the integral on the right hand side is $n!$ and in fact it is the canonical way to extend the factorial function to non-integers to find e.g. that $(-1/2)! = \sqrt{\pi},$ though of course we only need the integers here. After cancelling that through we find out that in fact, $$\hat Q = \pi,$$ or in other words we recover this property of resolving the identity even though not all of these functions are orthogonal, because the way that they're non-orthogonal just comes down to a constant multiplicative factor. We can therefore state unequivocally, $$1 = \iint dx~dy~\frac1\pi~|x,y\rangle\langle x,y|.$$ Your expression absorbs a $1/\sqrt{\pi}$ term into each of these kets, and writes $\pi^{-1/2} |x, y\rangle = |\alpha\rangle$ (where $\alpha = x + i y$) for short, both of which help in writing these expansions. One then finds similarly to the above expression with $A_{mn}$, that $$\langle A \rangle = \iint d^2\alpha~d^2\beta~\psi^*(\alpha)~A(\alpha,\beta)~\psi(\beta).$$The only cost to this notation is that we then have to express the above integrals with the more clumsy $\int d^2\alpha$ which is short for something like $d\alpha_x~d\alpha_y$ where $\alpha = \alpha_x + i \alpha_y.$
Quantum mechanics of a single particle
Quantum mechanics can be formulated in many ways. Let's review 2.
1. The Schrodinger picture
Position $x$ and momentum $p$ are time-independent operators, and the wavefunction $\psi(x, t)$ is a complex valued function of the particle's position and of time. The dynamics is defined via the Schrodinger equation
\begin{equation}
i\hbar \frac{\partial \psi}{\partial t} = H(x, p=-i\hbar \nabla) \psi
\end{equation}.
2. The Heisenberg picture
Position $x(t)$ and momentum $p(t)$ are time dependent operators, and the state is time independent (you can think of the state as defining the initial conditions). The dynamics are described by Heisenberg's equations
\begin{eqnarray}
\frac{dx}{dt} &=& \frac{i}{\hbar} [H, x] \\
\frac{dp}{dt} &=& \frac{i}{\hbar} [H, p]
\end{eqnarray}
This mimics Hamilton's equations from classical mechanics when written in terms of the Poisson bracket (here I'll use the subscript ${\rm cl}$ to denote that in this equation $x_{\rm cl}$ and $p_{\rm cl}$ are classical functions, not operators)
\begin{eqnarray}
\frac{dx_{\rm cl}}{dt} &=& \{H_{\rm cl}, x_{\rm cl}\} \\
\frac{dp_{\rm cl}}{dt} &=& \{H_{\rm cl}, p_{\rm cl}\}
\end{eqnarray}
Quantum field theory
Our goal is now to describe relativistic quantum mechanics. Let's first explain an incorrect approach, then describe two correct ways to do it.
0. An incorrect approach
The Schrodinger equation looks non-relativistic; it has one time derivative on the left hand side, and two spatial derivatives on the right hand side. So let's guess that the right way to describe a relativistic quantum particle is to generalize the Schrodinger equation. This was the original idea that led to the Klein-Gordon equation (describing a free particle)
\begin{equation}
-\frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} + \nabla^2 \psi = 0
\end{equation}
It turns out that this equation is relevant for relativistic quantum field theory, however this equation is not a generalization of the Schrodinger equation. This is the main point that the textbook authors want you to realize.
In fact, the entire notion of creating relativistic quantum mechanics is doomed from the start, because a relativistic quantum theory is necessarily one where the particle number can change. Tong's lecture notes, for example, give a physical argument for this; due to the uncertainty principle, if you localize a particle well enough it will have relativistic momentum and there is enough energy to transition to states where particle-anti-particle pairs are created.
1. The Schrodinger picture
The Schrodinger picture is rarely used in quantum field theory, but it can be formulated.
The wavefunction becomes a wavefunctional. It is still a function of all the relevant degrees of freedom of the system. However, instead of the degrees of freedom being the position of a single particle, the degrees of freedom are the values of a field $\Phi(x)$ at every location $x$. Therefore, the wavefunctional is a complicated object, $\psi[\Phi(x), t]$, which assigns a single probability amplitude for every possible field configuration $\Phi(x)$. (This usually takes students some thinking before they fully get it).
With this properly generalized and much more complicated wavefunctional, the Schrodinger equation actually looks essentially the same:
\begin{equation}
i \hbar \frac{\partial \psi}{\partial t} = H \psi
\end{equation}
except now the Hamiltonian is a functional of the field operator $\Phi(x)$ and its conjugate momentum $\pi(x)$.
2. The Heisenberg picture
This approach is much more common, and we will finally see the Klein-Gordon equation appearing in the correct place.
The field operators obey the Heisenberg equations of motion
\begin{eqnarray}
\frac{\partial \Phi(x,t)}{\partial t} &=& [H, \Phi] \\
\frac{\partial \pi(x,t)}{\partial t} &=& [H, \pi]
\end{eqnarray}
For the case of a free field, the Hamiltonian is
\begin{equation}
H = \int d^3 x \frac{1}{c^2} \frac{1}{2} \pi^2(x) + \frac{1}{2} (\nabla \Phi)^2
\end{equation}
You can work out all these commutators and you will find that $\Phi$ obeys the Klein-Gordon equation
\begin{equation}
-\frac{1}{c^2} \frac{\partial^2\Phi}{\partial t^2} + \nabla^2 \Phi = 0
\end{equation}
Here we see that the same equation appears that we might have guessed by generalizing the one particle Schrodinger equation. But the correct interpretation is not that $\Phi$ is a wavefunction of a particle. The interpretation of this equation is that it describes the evolution of the operator $\Phi(x, t)$; it is properly understood as a generalization of the Heisenberg equation, not the Schrodinger equation. It is analogous to the corresponding equation of a classical field, much like how Heisenberg's equations in quantum mechanics are analogous to the classical Hamiltonian equations for a single particle.
Best Answer
No, the words don't have any significantly different meaning than usual. From dictionary dot com: "Overlap: ... extend over and cover a part of..."
The integral is not just the product, it is the integration of the product.
Sakurai is basically saying that the "overlaps" are the regions where both the functions are not zero. If, say, one of the functions was only non-zero for $x>10$ and one of the functions was only non-zero for $x<-572$, then there would be no "overlap" and the integral would be zero. (Of course, there are other ways for the total integral to be zero, but let's not worry about that now.)
No, there is not really any "deeper" meaning. If there is any deeper meaning, it is in the meaning of the "bra." Ultimately, a "bra" (an element of the "dual space") is just a linear function of a "ket" (an element of the Hilbert space). Sakurai is just saying that since we have this nice completeness relation, we can use it as a concrete and effective way of calculating functions like $\langle \beta|\alpha\rangle$.