Let $\mathcal{M}$ be a manifold. Let $d : \mathcal{M} \times \mathcal{M} \rightarrow \mathbb{R}_{\geq 0}$ be the geodesic distance defined by $$d(x,y) = \inf \left\{ \int_0^1 \|\gamma'(t)\| \,\mathrm{d}t \;\middle|\; \gamma : (0, 1) \rightarrow \mathcal{M}, \gamma(0) = x, \gamma(1) = y \right\}$$
That is, $d(x,y)$ is the minimum length of any curve that connects $x$ and $y$. Let $\gamma_1, \gamma_2 : (0, 1) \rightarrow \mathcal{M}$ be two curves on $\mathcal{M}$. Suppose that \begin{align*} \gamma_1(0) &= \gamma_2(0) \\ \gamma_1'(0) &= \gamma_2'(0) \end{align*}
What is the tighest upper bound on $d(\gamma_1(t), \gamma_2(t))$ in terms of $|t|$?
I'm not sure what you're hoping for here. But without further hypotheses, there is no universal upper bound, since $\gamma_1$ and $\gamma_2$ could have arbitrarily large speeds and head off in different directions.
But if you assume that $|\gamma_1'(t)|$ and $|\gamma_2'(t)|$ are both bounded above by a constant $c$, then an easy computation shows that $d(\gamma_1(t),\gamma_2(t))\le 2 c |t|$. This is sharp, because it's possible to construct constant-speed curves in $\mathbb R^2$ that start at the origin with the same initial velocity, and then very quickly veer sharply in opposite directions, so that the distance is as close as you want to $2c|t|$.
One other thing that can be said is that for any given $\gamma_1,\gamma_2$, there's a constant $C$ such that $d(\gamma_1(t),\gamma_2(t))\le C t^2$ for small $t$ (this is essentially just Taylor's theorem in local coordinates); but there's no universal upper bound on $C$ because it would contradict the argument in the previous paragraph that you can get as close as you want to $2c|t|$.