References to understand $\frac{d}{dt} d(x,y) = \frac 12 \inf \int_\gamma g'(S,S) ds$, for flows on manifolds

In general, $d_t(x, y)$ is not differentiable in $t$. Let's assume that $x, y$ are so chosen so that it is differentiable at $t=0$.

Let $\gamma$ be any minimizing geodesic in $(M, g_0)$ so that $\gamma(0)=x$, $\gamma(1) = y$. Then

\begin{align} \frac{d}{dt} \bigg|_{t=0} L_{g_t} (\gamma)&= \frac{d}{dt}\bigg|_{t=0} \int_0^1 g_t( \gamma'(u) , \gamma'(u)) du \\ &= \frac 12 \int_0^1 \frac{\dot{g}_0 (\gamma'(u), \gamma'(u))}{\sqrt{g_0(\gamma'(u), \gamma'(u))}} du \\ &= \frac 12 \int_{\gamma} \dot g_0 (S, S) ds. \end{align} Since $d_0(x, y) = L_{g_0} (\gamma)$ and $d_t(x, y) \le L_{g_t}(\gamma)$, for $t>0$ we have

$$ \frac{d_t(x, y) - d_0(x, y)}{t} \le \frac{L_{g_t} (\gamma) - L_{g_0}(\gamma)}{t}$$

Taking $t\to 0^+$ on both sides, $$ \frac{d}{dt}\bigg|_{t=0} d_t(x,y) \le \frac 12 \int_{\gamma} \dot g_0 (S, S) ds.$$

On the other hand, for $t <0$,

$$ \frac{d_t(x, y) - d_0(x, y)}{t} \ge \frac{L_{g_t} (\gamma) - L_{g_0}(\gamma)}{t}$$

Taking $t\to 0^-$,

$$ \frac{d}{dt}\bigg|_{t=0} d_t(x,y) \ge \frac 12 \int_{\gamma} \dot g_0 (S, S) ds.$$

Thus we have

$$ \frac{d}{dt}\bigg|_{t=0} d_t(x,y) = \frac 12 \int_{\gamma} \dot g_0 (S, S) ds$$ if $d_t(x, y)$ is differentiable at $t=0$.

Since this is true for all such minimizing geodesics, we can take the infimum among all such geodesics (Obviously we have shown something stronger, that the equality holds for each minimizing geodesic $\gamma$ with the need to take infimum).

Remark Just for fun, I am including a simple example where $d_t(x, y)$ is not differentiable at $t=0$. Let $M = \mathbb R/\mathbb Z$, $x = 1/4$, $y = 3/4$ and let $b$ be a nonnegative, nonzero bump function supported in $[1/3, 2/3]$ and let $g_t = 1 + t b$. At $t=0$, $x, y$ is joined by two geodesics in $(M, g_0)$, $\gamma_1(t) = 1/4+ t/2$, and $\gamma_2(t) =1/4 - t/2$. When $t<0$, $\gamma_1$ is shorter than $\gamma$ and when $t>0$, $\gamma_2$ is shorter. Thus

$$ d_t(x, y) = \begin{cases} \frac 12 - t\int_0^1 b & \text{ when } t<0, \\ \frac 12 & \text{ when } t\ge 0 \end{cases}$$ and $d_t(x, y)$ is not differentiable at $t=0$.