I think, the gauge transformation has in its definition the superspace $y_\mu$, which is equal with $x_\mu + i \theta \sigma_\mu \bar \theta$. Thus in its Taylor expansion appears only 3 terms: $Ω(y) = Ω(x) + i \theta \sigma_\mu \bar \theta Ω(x) + \frac{1}{4} \theta \theta \bar \theta \bar \theta \Box{Ω(x)}$.

I think, the gauge transformation has in its definition the superspace $y_\mu$, which is equal with $x_\mu + i \theta \sigma_\mu \bar \theta$. Thus in its Taylor expansion appears only 3 terms: $Ω(y) = Ω(x) + i \theta \sigma_\mu \bar \theta \partial_\mu Ω(x) + \frac{1}{4} \theta \theta \bar \theta \bar \theta \Box{Ω(x)}$.

I think, the gauge transformation has in its definition the superspace y_\mu, which is equal with x_\mu + i \theta \sigma_\mu \bar \theta. Thus in its Taylor expansion appears only 3 terms: Ω(y) = Ω(x) + i \theta \sigma_\mu \bar \theta Ω(x) + \frac{1}{4} \theta \theta \bar \theta \bar \theta \Box Ω(x).

I think, the gauge transformation has in its definition the superspace $y_\mu$, which is equal with $x_\mu + i \theta \sigma_\mu \bar \theta$. Thus in its Taylor expansion appears only 3 terms: $Ω(y) = Ω(x) + i \theta \sigma_\mu \bar \theta Ω(x) + \frac{1}{4} \theta \theta \bar \theta \bar \theta \Box{Ω(x)}$.