Let $(M,g)$ be a Riemannian manifold, and let $p \in M$.
Is it true that for sufficiently small $r>0$, for any $v \in T_pM$ with norm $|v|_g=r$, $$ d_g(\exp_p(v),\exp_p(-v))=2r $$ holds?
I have an argument (below) that shows that this holds for at least one such $v$ with norm $r$.
We claerly have $d_g(\exp_p(v),\exp_p(-v)) \le2r$ by triangle inequality.
I thought using the existence of convex neighbourhoods but so far without luck.
Here is an almost(?) rigorous argument for the existence of such a $v$;
Let $S_r(p) \subseteq M$ be the sphere of radius $r$ around $p$.
Let $q=\exp_p(w),q'=\exp_p(w')$ be points such that $$ \text{diam}(S_r(p))=d_g(q,q'). $$
Let $\gamma$ be a minimizing geodesic between $q,q'$. Then $\gamma$ is orthogonal to $S_r(p)$ at $q,q'$, i.e. if $\gamma(0)=q, \gamma(L)=q'$, then $$ \dot \gamma(0) \perp T_qS_r(p), \,\,\,\,\dot \gamma(L) \perp T_{q'}S_r(p). $$ This implies that $\dot \gamma(0)$ is in the direction of the unique geodesic from $q$ to $p$ (the "backwards" of the radial geodesic $p \to q$), and similarly, $\dot \gamma(L)$ is in the direction of the unique geodesic from $q'$ to $p$.
So, $\gamma$ passes through $p$, and we have $$ d_g(q,q')=d_g(q,p)+d_g(p,q')=2r $$ as required.
On a second thought, I think that the result follows from the existence of what is called "totally normal neighbourhoods".
Specifically, there exists an open set $W$ containing $p$ and $\delta>0$, such that for every $q \in W$, $\exp_q$ is a diffeomorphism on $B_{\delta}(0) \subset T_qM$, and $W \subseteq \exp_q(B_{\delta}(0))$.
Now, take $0<r< \delta/2$, and let $q=\exp_p(v),q'=\exp_p(-v)\in S_r(p) \subset W$. Then $q'=\exp_q(w)$ for a unique $w \in B_{\delta}(0) \subset T_qM$, and $$d_g(q,q')=|w|.$$
However, we already have a candidate for $w$:
If $\gamma:[0,1] \to M$ is the geodesic from $p$ to $q$, i.e. $\gamma(t)=\exp_p(tv)$, then $w=-2\dot \gamma (1)$. So, $$|w|=2|\dot \gamma (1)|=2|\dot \gamma (0)|=2|v|=2r.$$
Thus $d_g(q,q')=2r$ as required.