= Solution
Introduce $y^\mu=x^\mu+i\theta\sigma^\mu\bar\theta$ and retain the same <Grassmann variables>. The <supersymmetric derivatives in chiral coordinates> follow from the <left Grassmann derivative> chain rule. Differentiating $y$ gives
$$
\partial_\alpha y^\mu=i(\sigma^\mu\bar\theta)_\alpha,\qquad
\bar\partial_{\dot\alpha}y^\mu=-i(\theta\sigma^\mu)_{\dot\alpha}.
$$
The minus sign in the second relation comes from moving the odd <Grassmann derivative> past $\theta$. Hence, as operators on a <superfield> expressed in $(y,\theta,\bar\theta)$,
$$
\left.\partial_\alpha\right|_x=\left.\partial_\alpha\right|_y+i(\sigma^\mu\bar\theta)_\alpha\partial_{y^\mu},\qquad
\left.\bar\partial_{\dot\alpha}\right|_x=\left.\bar\partial_{\dot\alpha}\right|_y-i(\theta\sigma^\mu)_{\dot\alpha}\partial_{y^\mu},
\qquad \partial_{x^\mu}=\partial_{y^\mu}.
$$
Substitution into the two <supersymmetric covariant derivatives> adds the two unbarred spacetime terms and cancels the two barred ones:
$$
\boxed{D_\alpha=\left.\partial_\alpha\right|_y+2i(\sigma^\mu\bar\theta)_\alpha\partial_{y^\mu},\qquad
\bar D_{\dot\alpha}=-\left.\bar\partial_{\dot\alpha}\right|_y.}
$$
These operator equalities prove both requested actions on $V$. In particular, a <chiral superfield> becomes independent of $\bar\theta$ at fixed $y$. The TeX aid corrupts the second formula by replacing its ordinary barred derivative with a covariant one; the original PDF has the ordinary $-\bar\partial_{\dot\alpha}$.
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