= Solution
A <scalar superfield transformation> is a pullback under a <supertranslation>. Since that transformation mixes $x$ and $\theta$, an odd-coordinate derivative acquires an extra spacetime-derivative term under the coordinate Jacobian. It therefore does not transform as a spinor <superfield> merely by differentiating the scalar transformation law. This is <partial derivatives and superfield covariance>.
For explicit signs, split the <Majorana spinor> coordinates into $\theta^\alpha,\bar\theta^{\dot\alpha}$ and use <left Grassmann derivatives>. Choose the differential generators
$$
Q_\alpha=\partial_\alpha-i\sigma^m_{\alpha\dot\beta}\bar\theta^{\dot\beta}\partial_m,
\qquad
\bar Q_{\dot\alpha}=-\bar\partial_{\dot\alpha}
+i\theta^\beta\sigma^m_{\beta\dot\alpha}\partial_m.
$$
For example, $\{\partial_\alpha,\bar Q_{\dot\beta}\}=i\sigma^m_{\alpha\dot\beta}\partial_m$, which is nonzero. Commuting the even <supersymmetry> variation through $\partial_\alpha$ consequently gives the unwanted translation term.
Replace the ordinary odd-coordinate derivatives by the <supersymmetric covariant derivatives>
$$
\boxed{D_\alpha=\partial_\alpha+i\sigma^m_{\alpha\dot\beta}\bar\theta^{\dot\beta}\partial_m,
\qquad
\bar D_{\dot\alpha}=-\bar\partial_{\dot\alpha}
-i\theta^\beta\sigma^m_{\beta\dot\alpha}\partial_m.}
$$
The graded product rule cancels the mixed coordinate/derivative terms, giving $\{D,Q\}=\{D,\bar Q\}=\{\bar D,Q\}=\{\bar D,\bar Q\}=0$. Thus $D_\alpha V$ and $\bar D_{\dot\alpha}V$ transform as spinor <superfields>. Their own mixed algebra is $\{D_\alpha,\bar D_{\dot\beta}\}=-2i\sigma^m_{\alpha\dot\beta}\partial_m$, with equal-chirality <anticommutators> zero. This covariant differentiation is also what makes a <chiral superfield> constraint $\bar D_{\dot\alpha}Z=0$ <supersymmetry> invariant. Negating every barred derivative is another convention, but changes the displayed mixed-algebra sign.
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