Analytic formula and an exponential bound
For $r \in [0, 1]$, define $\tau_m(r) \in [0, 1]$ by $$ \tau_m(r) := \frac{1}{\lambda(B_m(r))}\int_{B_m(r)} \frac{\lambda(B_m(1) \cap (B_m(1) + x))}{\lambda(B_m(1))}dx. $$
It has been proven by other users that
$$ \begin{split} \tau_m(r) &= \frac{1}{V_m^{cap}(r, 0)}\int_{0}^r ms^{m-1}V_m^{cap}(1, s/2)ds = \frac{1}{r^mV_m^{cap}(1, 0)}\int_{0}^r ms^{m-1}V_m^{cap}(1, s/2)ds\\ &= \frac{1}{V_m^{cap}(1, 0)}\int_{0}^1 ms^{m-1}V_m^{cap}(1, rs/2)ds, \end{split} $$
where $V_m(r; h)$ is the volume (i.e Lebesgue measure) of the half-lens $\{x \in B_m(r) \mid x_1 \ge h\}$. I propose to do the actual calculations and get an analytic formula in terms of special functions (beta, gamma, etc.).
The end result will be the following result on the exponential decay of $\tau_m(r)$ as a function on the dimension $m$.
Theorem. For every $r \in [0, 1]$ and large integer $m$, we have the bound $$ \tau_m(r) \le \exp\left(-\frac{mr^2}{8-r^2}\right) + \frac{r(1-(r/2)^2)^{m/2}}{\sqrt{m\pi}}. $$
Proof.
Now, it is a classical computaiton that $V_m^{cap}(1, h) = \beta_{1-h^2}\left(\frac{m+1}{2},\frac{1}{2}\right)$, where $\beta_x(a, b) := \int_0^xt^{a-1}(1-t)^{b-1}dt$ defines the incomplete beta function. In particular, $V_m^{cap}(0) = \beta_{1}\left(\frac{m+1}{2},\frac{1}{2}\right)$.
Thus, integrating by parts, we get
$$ \begin{split} V_m^{cap}(1,0)\tau_m(1) &= \int_{0}^1 s^{m-1}V_m^{cap}(rs/2)ds = \int_{0}^1 \beta_{1-(rs)^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)ds^m\\ &= \left[s^m\beta_{1-(rs)^2/4}\left(\frac{m+1}{2},\frac{1}{2}\right)\right]_0^1 + r\int_0^1 s^m(1-(rs)^2/4)^{\frac{m-1}{2}}ds \\ &= \beta_{1-(r/2)^2}\left(\frac{m+1}{2},\frac{1}{2}\right) + R_m, \end{split} \tag{1} $$ where $R_m := r\int_0^1 s^m(1-(rs/2)^2)^{\frac{m-1}{2}}ds$. Now, consider the change of variable $s = 2\sqrt{t}$. This gives $s^m(1-s^2/4)^{(m-1)/2}ds=2^mt^{n/2}\cdot(1-t)^{(m-1)/2}\cdot s^{-1/2}ds = 2^mt^{(m-1)/2}(1-t)^{(m-1)/2}$. Thus,
$$ \begin{split} R_m &= r\int_0^1 s^m(1-(rs)^2/4)^{\frac{m-1}{2}}ds = \frac{2^m}{r^m}\int_0^{(r/2)^2}t^{(m-1)/2}(1-r)^{(m-1)/2}dt\\ &=: \frac{2^m}{r^m}\beta_{(r/2)^2}\left(\frac{m+1}{2},\frac{1}{2}\right). \end{split} \tag{2} $$
Combining (1) and (2) then gives the analytic formula,
Analytic formula. For any $r \in [0, 1]$ and positive integer $m$, it holds that $$ \tau_m(r) = I_{1-(r/2)^2}\left(\frac{m+1}{2},\frac{1}{2}\right) + \frac{(2/r)^m\beta_{(r/2)^2}\left(\frac{m+1}{2},\frac{m+1}{2}\right)}{\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)} $$
Bounding the first term. Let $\delta := (r/2)^2$ and $p = I_{1-\delta}((m+1)/2,1/2)$. It is known since Temme (1992)that for $p \in (0, 1)$ and large $a > 0$, the solution of the equation $p = I(t; a,b)$ is given (approximately) by
$$ t=t_p(a, b) \approx e^{-(1/a)Q_{1-p}(\Gamma(b,1))}, \tag{4} $$
where $Q_{1-p}(\Gamma(b,1))$ is the $1-p$ quantile of the unit-scale gamma distribution with shape parameter $b$. Now by standard concentration results (e.g see Boucheron et al. textbook),
$$ Q_{1-p}(\Gamma(b,1)) \le \log(1/p) + \sqrt{2b\log(1/p)}. \tag{5} $$
In particular, for $a=(m+1)/2$ and $b=1/2$ we get
$$ Q_{1-p}(\Gamma(1/2,1)) \le \log(1/p) + \sqrt{\log(1/p)} \le 2\log(1/p). \tag{6} $$
Putting (2), (4), and (6) together and using the basic inequality $1-e^{-z} \ge 2z/(2+z)\;\forall z \ge 0$, we see that
$$ \begin{split} \delta &\le 1 - t_{p}\left((m+1)/2,1/2\right) \ge 1-e^{-\frac{2Q_{1-p}(\Gamma(1/2,1))}{m+1}} \ge 1-e^{-\frac{2}{m+1}\left(\log\left(\frac{1}{p}\right) + \sqrt{\log\left(\frac{1}{p}\right)}\right)} \le \frac{2\alpha}{2+\alpha}, \end{split} $$
where $\alpha = \frac{2}{m+1}\log(1/p)$. Thus, $\alpha \ge \delta/(1 - \delta/2)$, from which
$$ \begin{split} I_{1-(r/2)^2}((m+1)/2,1/2) &= I_{1-\delta}((m+1)/2,1/2) = p = e^{-\frac{m+1}{2}\alpha} \le e^{-\frac{(m+1)\delta}{2-\delta}}\\ &\le e^{-\frac{m\delta}{2-\delta}} = e^{-\frac{mr^2}{8-r^2}} \end{split} $$
Bounding the second term. If $z \mapsto {}_2F_1(a, b, c; z)$ is the hypergeometric function, then one computes
$$ \begin{split} (2/r)^m\beta_{(r/2)^2}\left(\frac{m+1}{2},\frac{m+1}{2}\right) &= \frac{r}{m+1}{}_2F_1\left(\frac{m+1}{2},\frac{1-m}{2},\frac{m+3}{2},(r/2)^2\right) \\ &= \frac{r}{m+1}{}_2F_1\left(\frac{1-m}{2},\frac{m+1}{2},\frac{m+3}{2},(r/2)^2\right)\\ &\approx \frac{r}{m+1}{}_2F_1\left(\frac{1-m}{2},\frac{m+1}{2},\frac{m+1}{2},(r/2)^2\right)\\ &= \frac{r}{m+1}(1-(r/2)^2)^{(m-1)/2} \end{split} $$
On the other hand, Stirling's formula gives
$$ \beta_1\left(\frac{m+1}{2},\frac{1}{2}\right) \approx \Gamma(1/2)\left(\frac{m+1}{2}\right)^{-1/2} = \sqrt{\frac{\pi}{m+1}}. $$
Thus,
$$ \frac{2/r)^m\beta_{(r/2)^2}\left(\frac{m+1}{2},\frac{m+1}{2}\right)}{\beta_1\left(\frac{m+1}{2},\frac{1}{2}\right)} \approx \frac{r(1-(r/2)^2)^{(m-1)/2}}{\sqrt{\pi(m+1)}} $$
Putting every together then completes the proof of the theorem. $\quad\quad\quad\Box$