Let $x$ and $y$ are nonzero vectors in $\mathbb R^n$, and $\hat{x} := x/\|x\|$, $\hat{y}:=y/\|y\|$ be corresponding unit vectors. One can easily show that
$$ \|\hat{x}-\hat{y}\| \le \frac{\|x-y\|}{\min(\|x\|,\|y\|)}.\tag{1} $$
Indeed, w.l.og, suppose $\|x\| \le \|y\|$ and let $\alpha := \|y\|/\|x\| \ge 1$. Then, one computes $$ \|x-y\| = \|x\|\|\hat{x}-\alpha \hat{y}\| \ge \|x\|\|\hat{x}-\hat{y}\|, $$ where we've used the fact that if $u$ and $v$ are unit vectors and $\alpha \ge 1$, then $\|u-\alpha v\| \ge \|u-v\|$.
Question. Can we get an lower-bound for $\|\hat{x}-\hat{y}\|$ in as an increasing function of $\|x-y\|$ (and perhaps some dependence on $\|x\|$, and $\|y\|$) ?
Changing notation, let $u=\hat x$ and $v\hat y$ be unit vectors, and let $\alpha=\|x\|$ and $\beta=\|y\|$, so that $x=\alpha u$ and $y=\beta v$. Then \begin{align*} \|x-y\|^2 &= \|\alpha u-\beta v\|^2 = \alpha^2+\beta^2-2\alpha\beta(u\cdot v) \\ \|\hat x-\hat y\|^2 &= \|u-v\|^2 = 2-2(u\cdot v), \end{align*} and these two equations can be solved exactly to yield $$ \|\hat x-\hat y\| = \sqrt{\frac{\|x-y\|^2-(\alpha-\beta)^2}{\alpha\beta}} = \sqrt{\frac{\|x-y\|^2-(\|x\|-\|y\|)^2}{\|x\|\cdot\|y\|}}, $$ which is indeed an increasing function of $\|x-y\|$ with dependence on $\|x\|$ and $\|y\|$. (Note that it quickly implies the upper bound given in the OP as well.)