I have two +q on $(y_0/2,y_0 \cos \omega t)$ and $(-y_0/2,-y_0 \cos \omega t)$ , two -q on $(-y_0/2,y_0 \cos \omega t)$ and $(y_0/2,-y_0 \cos \omega t)$

Given $\vec E={ q\over4\pi \epsilon_0 c}\{{1\over (1-\vec \beta \cdot \hat n)^3R} \hat n \times [(\hat n - \vec \beta \times \dot {\vec \beta})]\}_{t_R}$ Find the E-field observed at $R$ from the origin, assume $R \gt \gt y_0$ and $\omega y/c \lt \lt 1$

I make the approximation as follow:

1.Distance from each point charge to the observer point is equal, say $R$

2.$\hat n = \cos \theta \hat x + \sin \theta \hat y$, $\vec \beta =\mp \omega y_0 \sin \omega t$, ${1\over (1-\vec \beta \cdot \hat n )^3}=1 \mp {3\over c }\omega y_0 \sin \omega t \sin \theta$, $\hat n \times (\hat n \times \vec \beta)=\mp {\omega^2 \over c}y_0 \cos \omega t \cos \theta \sin \theta \pm {\omega ^2 \over c} y_0 \sin \omega t \sin \theta \hat y$

When I substitute all this, I got $0$, where do I do wrong?

It is incorrect to assume that all charges are at the same distance from the observation point. You have to expand the distance as a Taylor series in $y/R$ and keep the highest order term in the end.