I) Let us for simplicity work in 1D with $\hbar=1$. (The generalization to higher dimensions is straightforward.) Moreover, let us for simplicity take an operator $\hat{f}(\hat{X},\hat{P})$ without any ordering ambiguities, i.e., each monomial term in the symbol $f(x,p)$ depends only on either $x$ or $p$, but not on both. Then one possible motivation of Wigner's phase space distribution
$$
\tag{1} w(x,p)~:=~ \int_{\mathbb{R}}\! {dy\over2\pi}e^{ipy}\psi^{*}(x+\frac{y}{2}) \psi(x-\frac{y}{2})
$$
goes as follows. The expectation value of the operator $\hat{f}(\hat{X},\hat{P})$ in the Schrödinger position representation
$$\tag{2} \hat{X} ~\longrightarrow~x, \qquad \hat{P} ~\longrightarrow~ -i \frac{\partial}{\partial x}, $$
reads
$$\langle\psi| \hat{f}(\hat{X},\hat{P})|\psi \rangle ~\stackrel{(2)}{=}~ \int_{\mathbb{R}} \!dx~ \psi^{*}(x) f\left(x, -i \frac{\partial}{\partial x}\right) \psi(x) \qquad\qquad $$
$$ ~\stackrel{\begin{matrix}\text{substitute}\\ x=x^{\pm}\end{matrix}}{=}~ \int_{\mathbb{R}^{2}}\! dx^{+}dx^{-}~ \delta(x^{+}-x^{-})\psi^{*}(x^{+}) f\left(\frac{x^{+}+x^{-}}{2}, -i \frac{\partial}{\partial x^{-}}\right) \psi(x^{-}) $$
$$ ~\stackrel{\delta\text{-fct}}{=}~ \int_{\mathbb{R}^{3}}\! {dx^{+}dx^{-}dp\over2\pi}~ e^{ip(x^{+}-x^{-})}\psi^{*}(x^{+}) f\left(\frac{x^{+}+x^{-}}{2}, -i \frac{\partial}{\partial x^{-}}\right) \psi(x^{-}) $$
$$ ~\stackrel{\text{int. by parts}}{=}~\int_{\mathbb{R}^{3}}\! {dx^{+}dx^{-}dp\over2\pi}~ e^{ip(x^{+}-x^{-})}\psi^{*}(x^{+}) f\left(\frac{x^{+}+x^{-}}{2}, p\right) \psi(x^{-}) $$
$$ ~\stackrel{\begin{matrix}\text{substitute}\\ x^{\pm} = x\pm \frac{y}{2}\end{matrix}}{=}~\int_{\mathbb{R}^{3}}\! {dx~dy~dp\over2\pi}~ e^{ipy}\psi^{*}(x+\frac{y}{2}) f(x, p) \psi(x-\frac{y}{2}) $$
$$ \tag{3} ~\stackrel{(1)}{=}~\int_{\mathbb{R}^{2}}\! dx~dp~w(x,p) f(x, p) .$$
That's the motivation!
II) For more general operators $\hat{f}(\hat{X},\hat{P})$,
we leave it for OP to show that if $f(x,p)$ is interpreted as the Weyl-symbol of the operator $\hat{f}(\hat{X},\hat{P})$, then the equation
$$\tag{3'} \langle\psi| \hat{f}(\hat{X},\hat{P}) |\psi\rangle ~=~\int_{\mathbb{R}^{2}}\! dx~dp~w(x,p) f(x, p)$$
continues to hold [at least for a sufficiently well-behaved function $f(x, p)$].
III) One important virtue, from a physics perspective, of the Weyl-ordering (as opposed to other ordering prescriptions) is that the operator $\hat{f}(\hat{X},\hat{P})$ formally becomes Hermitian for real functions $f:\mathbb{R}^2 \to\mathbb{R}$ and two Hermitian operators $\hat{X}$ and $\hat{P}$. Recall that Hermitian operators correspond to physical observables in quantum mechanics. For Weyl-ordering, see also e.g. this Phys.SE post.