$\newcommand{\ep}{\epsilon}$ $\newcommand{\al}{\alpha}$ $\newcommand{\be}{\beta}$
Let $M$ be a Riemannian manifold, $S \subset M,p \in M \setminus{S}$, and $v \in T_pM$ be a unit vector. Let $d_S$ be the distance function from $S$.
Suppose that for every unit speed path $\be(t)$ starting at $p$, there exists $\ep_{\beta}>0$ and a unit speed path $\al_{\beta}:(-\ep_{\beta},\ep_{\beta}) \to M$ also starting at $p$, such that $\dot \al_{\beta}(0) = v$ and $$ d_S(\al_{\beta}(t)) \le d_S(\be(t)) \quad \text{ for all } t \in (-\ep_{\beta},\ep_{\beta}). $$
Does there exist a single unit speed path $\al:(-\ep,\ep) \to M$ starting at $p$ with velocity $v$, that "beats" all the different $\beta$'s at once?
i.e. I want $\al(0)=p , \dot \al(0) = v,$ such that for every unit speed path $\be(t)$ starting at $p$, $$ d_S(\al(t)) \le d_S(\be(t)) \quad \text{ for all } t \in \left( -\tilde \ep_{\beta},\tilde \ep_{\beta} \right). $$
(I allow $\tilde \ep_{\beta} < \ep_{\beta}$.) Informally, I am assuming that $v$ is a "preferred direction of movement towards $S$ from $p$", and I ask if there is a single path in that direction, which beats all other paths for short times. My guess is that the geodesic $t \to \exp_p(tv)$ should do the job, but I am not sure how to prove that.
Assuming $M$ is complete, choose a sequence $x_i \in S$ for which $d(p,x_i)$ approaches $D = \text{inf}\{d(p,x) \mid x \in S\}$.
It follows that the set $Q = \{q \in \overline D \mid d(p,q)=D\}$ is not empty. The property of your vector $v$ that you state will be true if and only if $v$ is the initial direction of some unit speed geodesic $\alpha$ from $p$ to some $q \in Q$.
The point is that if $\alpha$ does have that form then $d_S(\alpha(t)) = D-t$, and $d_S(\beta(t)) \ge D-t$ for each of your $\beta$ paths.
If on the other hand if $\alpha$ does not have that form then you can take $\beta$ to be a path which does have that form.