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Let $A$ be a finite $K$-approximate group in $G$, let $H\le G$, and let $m\ge 2$ be an integer. Then $B=A^m\cap H$ is a $K^{2m-1}$-approximate group.

The higher product bound for an approximate group covers $A^{2m}$ by at most $K^{2m-1}$ left translates of $A$. In every translate meeting $B^2$, choose one representative $b\in B^2$. If $z$ is another point in that intersection, then $b^{-1}z\in A^{-1}A\cap H=A^2\cap H\subseteq B$. Thus the chosen representatives give $B^2\subseteq YB$ with $|Y|\le K^{2m-1}$. Identity containment and inversion symmetry are inherited from $A^m\cap H$.

The restriction $m\ge 2$ matters. In $\mathbb{Z}\times {\mathbb{Z}}_2$, put $A=([-N,N]\times \{1\})\cup (S\times \{0\})$, where $S\subseteq [-N,N]$ is any symmetric set containing zero. Four translates of $A$, with first-coordinate shifts $\pm N$ and both second-coordinate shifts, cover $[-2N,2N]\times {\mathbb{Z}}_2$, hence cover $A^2$. Thus $A$ is a $4$-approximate group, while $A\cap (\mathbb{Z}\times \{0\})=S\times \{0\}$ can have arbitrarily large doubling constant.

WriteUnder the improved bounds allowed in the question, $r=O({\log }^{O(1)}(2K))$ andMoreover $AC\subseteq A^{O(1)}$, whose size is at most $K^{O(1)}|A|$ by the higher product bound for an approximate group. ThusThe Ruzsa covering lemma supplies $X\subseteq A$ of size at most $\exp (O({\log }^{O(1)}(2K)))$ such thatTaking the $B_i$ to be the factors in this last product giveswhere $k\le O({\log }^{O(1)}(2K))$ and every $B_i$ is a $K^{O(1)}$-approximate group in $A^{O(1)}$ generating a subgroup of step less than $s$.