By the obvious scaling invariance we may assume $r=1$. Let $D=Im(f_{0})\cap Im(f_{1})$. Since $\epsilon$ is small, we may assume $\hat{A}(3,8)$ is contained in $f_{0}^{-1}(D)$. Then there is a constant $C_{1}$ independent of $\epsilon$ such that the map $\psi=10^{-1}f_{1}^{-1}\circ f_{0}:\hat{A}(3,8)\rightarrow\mathbb{R}^{4}$ satisfies $|\psi^{*}g_{0}-g_{0}|_{C^{4}}\leq C_{1}\epsilon$, and $(1-3\epsilon)|x|\leq|\psi(x)|\leq(1+3\epsilon)|x|$. By Lemma 5.8 there is an isometry $P$ of $\mathbb{R}^{4}$ such that $|P\circ\psi-Id|_{C^{3}}\leq K(C_{1}\epsilon)$ on $\hat{A}(4,7)$. We write $P(x)=R(x+\xi)$ for a rotation $R$ and a translation $\xi$. Then it is easy to see that $|R\circ\psi-Id|_{C^{3}}\leq C_{2}(\epsilon)$ with $\lim_{\epsilon\rightarrow 0}C_{2}(\epsilon)=0$, and $R(\hat{A}(1-\delta,100+\delta))$ contains $\hat{A}(1,100)$. Choose a cut-off function $\chi(x)$ on $\hat{A}(1,100)$ with $\chi(x)=1$ for $|x|\leq 5$ and $\chi(x)=0$ for $|x|\geq 6$. Using the map $f_{0}$ we get a corresponding cut-off function on $B(q,200)$, still denoted by $\chi$. Then $|\chi|_{C^{4}_{g}}\leq C_{3}$ for a constant $C_{3}$ independent of $\epsilon$. Clearly $\chi(p)=0$ when $p\notin Im(f_{1})$ and $\chi(p)=1$ when $p\notin Im(f_{0})$. Define $h:Imf_{0}\cup Imf_{1}\rightarrow\mathbb{R}^{4}$ sending $p$ to $10^{-1}\chi(p)R\circ f_{1}^{-1}(x)+(1-\chi(p))f_{0}^{-1}(x)$. Then for $\epsilon$ sufficiently small we have $h=f_{0}^{-1}$ on $A(8,100)$ and $h=10^{-1}R\circ f_{1}^{-1}$ on $A(10^{-1},3)$, and $|h^{*}g_{0}-g|_{C^{2}_{g}}\leq C(\epsilon)$ with $\lim_{\epsilon\rightarrow 0}C(\epsilon)=0$. Define $f=h^{-1}$. Then $f(x)$ meets the required properties. ∎