Let $(\mathcal M,g)$ be a connected Riemannian manifold embedded in $\Bbb R^K$.
Define the usual ball as $$ B_R(x) := \{y\in\Bbb R^K: |x-y|<R \} $$ for $x\in\Bbb R^K$.
For the ball in $M$, we write $$ B^{\mathcal M}_R(z) := \{ w\in \mathcal M:d(z,w)<R\} $$ for $z\in\mathcal M$, where $d(\cdot,\cdot)$ is the geodesic distance.
Is it always that case that, for a fixed $x_0\in\mathcal M$, we may find $R>0$ such that $$ B^{\mathcal M}_{R/2}(x_0) \subset B_R(x_0)\cap\mathcal M\subset B^{\mathcal M}_{2R}(x_0)\quad ? $$
What I am willing to assume is that $\mathcal M$ is compact. In this setting, can we find an $R$ that works for all $x\in\mathcal M$ uniformly as well?
I have a feeling that this should be true but I cannot find a reference to this. I would truly appreciate if anyone can give a nice argument or provide a reference to this result.
Since any path in $\mathcal M$ has the same length in $\mathbb R^K,$ we know $d(x,y) \le |x-y|$ and thus the inclusion $B_{R/2}^{\mathcal M}(x) \subset B_R (x) \cap \mathcal M$ always holds - we can even use $R$ instead of $R/2.$
The other inequality is where we need to "zoom in". Since $\mathcal M$ is an embedded submanifold, it is locally a graph: for small enough $R_0$ we can write $B_{R_0}(x_0)\cap \mathcal M$ as the graph of some function $$f : B_{R_0}(0) \subset \mathbb R^n \to \mathbb R^{K-n}$$ with $f(x_0) =0, Df(x_0)=0.$ (This usually involves changing coordinates on $\mathbb R^K$ by a rigid motion.)
The only thing stopping us from enlarging the domain on which such a graphical representation works is when $\mathcal M$ has a vertical tangent; so in the compact case any $R_0 \le 1/\sup |A|$ works uniformly. (Here $A$ is the second fundamental form, so $1/\sup |A|$ is a lower bound on radii of curvature.)
In such a coordinate patch, the second fundamental form $A$ is related to the Hessian of $f$ by $$D^2 f = \sqrt{1 + |Df|^2}A.$$ Since $Df = 0$ at the origin, we can estimate $\sqrt{1+|Df|^2} \le 1 + |Df| \le 1 + r\sup|D^2f|$ where $r$ is the horizontal distance to the origin; so for points closer than $r < 1/\sup |A|$ we have $$\sup|D^2 f| \le \frac{\sup|A|}{1-r\sup|A|}.$$
For any $Y=(y,f(y)) \in B_{R_0}(0)\cap \mathcal M$ we can then bound the intrinsic distance $d(0,Y)$ by the length of the path $(\gamma(t),f(\gamma(t)),$ where $\gamma$ is a straight line (in the horizontal $\mathbb R^n$) from the origin to $y.$ This arclength is simply $$L(\gamma) = \int_0^{|y|} \sqrt{1 + Df(\dot \gamma(t))^2} dt,$$ which we can bound using our estimate for $|D^2 f|$ as $$L(\gamma) \le \int_0^{|y|}1+|Df|\le |y|+ \sup|D^2f||y|^2\le\frac{|y|}{1-|y|\sup|A|}.$$ If we restrict to a ball of radius $R \le 1/(2 \sup |A|)$ then this yields $$d(0,Y) \le L(\gamma) \le 2|y|\le 2|x_0 - Y|,$$ so we have $B_R(x_0) \cap \mathcal M \subset B_{2R}^\mathcal M(x_0)$ as desired.