Let $(M,g)$ be a Riemannian manifold. Let $p\in M$. Is it possible to find a chart $\varphi: p\in U \to \mathbb R^d$ such that the geodesic distance $d_M$ on $U$ is close to the euclidean distance $d_E$ on $\phi_i(U_i)$?
More precisely, if I take $\phi_i = \exp^{-1}_p: \{x\in M: d_M(x,p)\le \delta\} \to \{v\in \mathbb R^d: d_E(v,0 )\le \delta\}$, can I find bounds $$A(p, M ,\delta) \le \frac{d_M(x,y)}{d_E(\exp^{-1}_p(x), \exp^{-1}_p(y))}\le B(p, M ,\delta)\enspace,$$ such that $A(p, M ,\delta)\small{\nearrow} \normalsize1$ and $B(p, M ,\delta)\small\searrow \normalsize1$ as $\delta\to 0$? Presumably, $A$ and $B$ will depend only on the curvature at $p$ and on $\delta$.
Of course, if $x=p$, then $A=B=1$ by the definition of the exponential map. The difficult part is $x,y\neq p$.
I think I know how to derive $B$. I would consider $\delta$ sufficiently small such that the geodesic ball is convex. A path in $\{v\in \mathbb R^d: d_E(v,0 )\le \delta\}$ from $\exp^{-1}_p(x)$ to $\exp^{-1}_p(y)$ induces a path $\gamma$ in $M$ of length
$$l(\gamma) = \int_0^1 \sqrt{g(\gamma'(t), \gamma'(t))} dt$$ with $g(\gamma', \gamma') = \Vert \gamma' \Vert^2 - \frac 1 3 R_{\mu \sigma \nu \tau}(p) ~ \gamma^\sigma \gamma^\tau \gamma'^\mu \gamma'^\nu+ \mathcal O(|\gamma|^3)$. So if $\gamma$ is a straight line, then $\gamma'(t) = \exp^{-1}_p(x) - \exp^{-1}_p(y)$ and $l(\gamma)$ depends directly on $d_E(\exp^{-1}_p(x), \exp^{-1}_q(x))$.
However, I don't know how to find the lower bound $A$.
Yes, the exponential map provides a diffeomorphism which is $(1+\epsilon)$-bi-Lipschitz on sufficiently small balls, for any $\epsilon>0$. This follows from the fact that the Jacobian of the exponential map at $p$ is the identity map.