[step:Use the Gauss lemma to eliminate radial-angular cross terms]
Define the polar coordinate domain
\begin{align*}
D_p:=\{(r,v)\in (0,\infty)\times S_pM:0<r<c(v)\}
\end{align*}
and the polar coordinate map
\begin{align*}
P:D_p&\to U\\
(r,v)&\mapsto \exp_p(rv).
\end{align*}
For $(r,v)\in D_p$, the tangent space splits as
\begin{align*}
T_{(r,v)}D_p = \mathbb{R}\partial_r \oplus T_vS_pM.
\end{align*}
The radial derivative is
\begin{align*}
dP_{(r,v)}(\partial_r)=\dot{\gamma}_v(r),
\end{align*}
so, since $\gamma_v$ is unit-speed,
\begin{align*}
g_{P(r,v)}\bigl(dP_{(r,v)}(\partial_r),dP_{(r,v)}(\partial_r)\bigr)=1.
\end{align*}
For $\xi\in T_vS_pM$, differentiating the map $(r,v)\mapsto rv$ in the angular direction gives the tangent vector $r\xi\in T_{rv}(T_pM)$, hence
\begin{align*}
dP_{(r,v)}(\xi)=d(\exp_p)_{rv}(r\xi).
\end{align*}
The Gauss lemma states that the differential of the exponential map sends radial and angular tangent directions to $g$-orthogonal tangent vectors. Therefore
\begin{align*}
g_{P(r,v)}\bigl(dP_{(r,v)}(\partial_r),dP_{(r,v)}(\xi)\bigr)=0
\end{align*}
for every $\xi\in T_vS_pM$.
Define, for each $r>0$ and $v\in S_pM$ with $r<c(v)$, the angular [bilinear form](/page/Bilinear%20Form)
\begin{align*}
(g_r)_v:T_vS_pM\times T_vS_pM&\to \mathbb{R}\\
(\xi,\eta)&\mapsto g_{\exp_p(rv)}\bigl(d(\exp_p)_{rv}(r\xi),d(\exp_p)_{rv}(r\eta)\bigr).
\end{align*}
Since $P$ is a diffeomorphism, $dP_{(r,v)}$ is injective; hence $(g_r)_v$ is positive definite on $T_vS_pM$. Smoothness of $g_r$ follows from smoothness of $g$, of $\exp_p$, and of the multiplication map $(r,v)\mapsto rv$.
Thus, with respect to the decomposition $\mathbb{R}\partial_r\oplus T_vS_pM$, the pulled-back metric satisfies
\begin{align*}
P^*g = dr\otimes dr + g_r.
\end{align*}
This is the asserted geodesic polar coordinate form. The Gauss lemma is used here as a background result not yet resolved to a wiki theorem ID.
[/step]