= Solution
Choose the given <short root> $\alpha$ as a <simple root>, and let $\beta$ be the long <simple root>. This entails no loss of generality: the <Weyl group> is transitive on the <short roots> of the <G2 root system>. The relevant <Cartan integers> are $\langle\beta,\alpha^\vee\rangle=-3$ and $\langle\alpha,\beta^\vee\rangle=-1$. The <positive roots> are
$$
\alpha,\ \beta,\ \alpha+\beta,\ 2\alpha+\beta,\ 3\alpha+\beta,\ 3\alpha+2\beta.
$$
Under $\mathfrak m_\alpha\cong\mathfrak{sl}_2$, the <root space> for $m\alpha+n\beta$ has $h_\alpha$ eigenvalue $2m-3n$. The subalgebra $\mathfrak m_\alpha$ itself is the three-dimensional irreducible $V(2)$ in the <Adjoint representation>.
The four <root spaces> along the <root string> $\beta,\alpha+\beta,2\alpha+\beta,3\alpha+\beta$ have eigenvalues $-3,-1,1,3$. Consecutive spaces are connected by nonzero <raising operators> and <lowering operators>, so their sum is $V(3)$. The negative <root string> supplies another $V(3)$. The <root spaces> for $\pm(3\alpha+2\beta)$ have eigenvalue zero, and neither adding nor subtracting $\alpha$ gives a root; they are two copies of $V(0)$. Finally, the one-dimensional space $\ker\alpha\subset\mathfrak t$ commutes with $\mathfrak m_\alpha$, giving one more $V(0)$. We have accounted for all $14$ dimensions, and therefore the <G2 adjoint branching to a short-root sl2 subalgebra> is
$$
\boxed{\mathfrak g\cong V(3)^{\oplus2}\oplus V(2)\oplus V(0)^{\oplus3}.}
$$
For a nonzero <root vector> $x\in\mathfrak g_\alpha$, its <Adjoint representation> action is a nonzero scalar multiple of the <raising operator> on each summand. On each irreducible $V(n)$ this operator has a one-dimensional <kernel>, including $V(0)$. Since the <Lie algebra centralizer> is that <kernel>,
$$
\boxed{\dim Z_{\mathfrak g}(x)=2+1+3=6.}
$$
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