# $U$ is an open set of $TM$, then there exists an $\epsilon$ such that $X =\{(x,v):r(x) < 2\epsilon, |v|_\bar g<2\epsilon\} \subset U$

Claim: Suppose that $M$ is a Riemannian manifold, $p\in M$ is a point and $(U,\varphi)$ is a normal coordinate chart around $p$. Let $\mathcal{U}$ be an open neighborhood of $(p,0)$ in $TM$, then there exists an $\epsilon$ such that $X =\{(x,v):r(x) < 2\epsilon, |v|_\bar g<2\epsilon\} \subset \mathcal{U}$. $\bar g$ is the corresponding Euclidean norm while $r(x)$ is the Euclidean radial distance with respect to the normal coordinate chart.

Why should this be true?

[Your original question was wrong: an open set in $TM$ may not contain a point of the form $(p,0)$ at all (and this point certainly is inside of such an $X$). I added to the question what I thought was needed. Hopefully this already clarified things up a bit.]
1. For any neighborhood $V$ of $p$ in $M$ there exists an $\varepsilon >0$ such that $$B=\{ x\in M | r(x) < \varepsilon \} \subseteq V$$ This is easy to deduce from the definitions of radial distance function and normal coordinate chart. Consider the open neighborhood $V\cap U$ of $p$. Map it through your coordinate (homeomorphic) function $\varphi \colon U \to \mathbb{R}^{n}$ to some open neighborhood $W$ of $\varphi(p)=0$ inside of $\mathbb{R}^{n}$. Then there exists an $\varepsilon >0$ such that $B_{\varepsilon }(0) \subseteq W$. By definition, for any $x\in U$, $r(x)$ is the Euclidean distance from the image through $\varphi$ of the point $x$ to the origin in $\mathbb{R}^{n}$ (the image of the point $p$). Therefore $$B=\varphi^{-1}(B_{\varepsilon }(0))\subseteq \varphi^{-1}(W)=V\cap U \subseteq V$$
2. For any neighborhood $V$ of $0$ in $T_{x}M$ there exists $\varepsilon >0$ such that $$A=\{ v\in T_{x}M | |v|_{\bar{g}} < \varepsilon \} \subseteq V$$ The idea is similar as before. You get a homeomorphism between $T_{\varphi(x) }\mathbb{R}^{n}\cong \mathbb{R}^{n}$ and $T_{x}M$ and you use it as before to get the same conclusion, using that the topology in $\mathbb{R}^{n}$ is induced by the euclidean norm.