Ok, here's an example: $$F:B^n_{1/(2e)}\subset\mathbb{R}^n\to\mathbb{R}^n,F(x)=-2\log(|x|)\cdot x,$$ where $B^n_{\epsilon}$ is the $n$-ball with radius $\epsilon$, and $F$ maps $B^n_{1/(2e)}$ homeomorphically onto its image. (We can extend $F$ smoothly to the whole $\mathbb{R}^n$ to make a homeomorphism.)
- $F$ is not differentiable at 0: assume it is, then the differential must be a scaling, say by $k\in\mathbb{R}$. Then $\lim_{x\to0}\frac{F(x)-kx}{|x|}=\lim_{x\to0}\frac{-2\log|x|-k}{|x|}\cdot x=0$, but $\lim_{x\to0}\frac{\log|x|}{|x|}=\infty$, a contradiction.
We next check that $F$ satisfies the property listed in the last paragraph of the question, with $\phi:S^{n-1}\to S^{n-1}$ being the identity. This implies that $(F,F):B^n_{1/2e}\times B^n_{1/2e}\to\mathbb{R}^n\times\mathbb{R}^n$ lifts to $\pi^{-1}(0,0)$ in the blow-up, which should suffice for the conclusion.
- Denote $e_1=\frac{\partial}{\partial x_1}\in S^{n-1}$ the unit vector parallel to the $x_1$-axis. Since our map $F$ commutes with rotations, it would suffice to check the conclusion for sequences of pairs of points $\{(x_i,y_i)\}_i$ such that $\frac{y_i-x_i}{|y_i-x_i|}=e_1$ -- if $\{(x_i,y_i)\}_i$ is such that $\lim_{i\to\infty}\frac{y_i-x_i}{|y_i-x_i|}=e_1$, then let $r_i$ be the rotation (with center 0) of $\mathbb{R}^n$ such that $r_i(y_i)-r_i(x_i)=r_i(y_i-x_i)$ is parallel to $e_1$; since $$r_i(F(y_i)-F(x_i))=r_i(F(y_i))-r_i(F(x_i))=F(r_i(y_i))-F(r_i(x_i)),$$ if the direction of the last term converges to $e_1$, then so is the direction of $F(y_i)-F(x_i)$, since $||r_i||\to0$ where $||r_i||$ is the $L^\infty$-norm of $r_i:S^{n-1}\to S^{n-1}$.
- Abusing notation we also write $e_1$ for the tangert vector in the $e_1$ direction at any point $x\in\mathbb{R}^n$. For $x\neq0$ let $s(x)=\frac{|d_xF(e_1)-\langle d_xF(e_1),e_1\rangle e_1|}{\langle d_xF(e_1),e_1\rangle}$ be the "slope" of $(d_xF)(e_1)$. We show $\lim_{x\to0}s(x)=0$: direct computation shows $$s(x)=\frac{2x_1\sqrt{x^2_2+\ldots+x^2_n}}{2x_1^2+|x|^2\log(|x|^2)};$$ so for $|x|$ small enough so that $\log|x|^2<-2$, $$|s(x)|=\frac{2|x_1|\sqrt{x_2^2+\ldots+x_n^2}}{|x|^2(-\log|x|^2)-2x_1^2}<\frac{2|x_1|\sqrt{x_2^2+\ldots+x_n^2}}{|x|^2(-\log|x|^2-2)}<\frac{1}{-\log|x|^2-2}\longrightarrow0\text{ as }x\to0,$$ where the last inequality follows from Cauchy's inequality.
- Let $\{(x_i,y_i)\in\mathbb{R}^n\times\mathbb{R}^n\}_{i=1}^\infty$ be such that $\frac{y_i-x_i}{|y_i-x_i|}=e_1, \lim_{i\to\infty}x_i,y_i=0$. Denote by $F_1:\mathbb{R}^n\to\mathbb{R}$ the first factor of $F$ and $G:\mathbb{R}^n\to\mathbb{R}^{n-1}$ the other factors. We show below that $$\lim_{i\to\infty}\frac{|G(y_i)-G(x_i)|}{|F_1(y_i)-F_1(x_i)|}=0$$ which would conclude the proof. Given $\epsilon>0$, let $\delta>0$ be such that $|s(x)|<\epsilon$ for all $x$ with $|x|<\delta$. We check $|G(y_i)-G(x_i)|<\epsilon\,|F_1(y_i)-F_1(x_i)|$ whenever $|x_i|,|y_i|<\delta$. Suppose the first coordinate of $x_i$ is smaller than that of $y_i$. Let $\gamma:[a,b]\to\mathbb{R}^n,\gamma(a)=x_i,\gamma(b)=y_i$ be the (unit speed) line segment running from $x_i$ to $y_i$ parallel to $e_1$. Since $(F_1\circ\gamma)'(t)>0$ for all $t$ and $s(\gamma(t))<\epsilon\implies|(G\circ\gamma)'(t)|<\epsilon|(F_1\circ\gamma)'(t)|$, $$|G(y_i)-G(x_i)|\leq\int_a^b|(G\circ\gamma)'(t)|dt<\epsilon\int_a^b|(F_1\circ\gamma)'(t)|dt=\epsilon\,|F_1(y_i)-F_1(x_i)|.$$