= Solution
A <horizontal lift> of $\gamma$ from $p$ is a curve $\widetilde\gamma$ in $E$ projecting to $\gamma$, starting at $p$, and tangent to the horizontal distribution of the connection. In a local frame write $\widetilde\gamma(t)=(\gamma(t),v(t))$. Horizontality is the linear ordinary differential equation
$$
v'(t)+A_{\gamma(t)}(\dot\gamma(t))v(t)=0,
$$
whose initial-value theorem gives local existence and uniqueness; successive trivializations continue the lift.
A <geodesic> satisfies $\nabla_{\dot\gamma}\dot\gamma=0$, and
$$
\exp_p(v)=\gamma_v(1)
$$
for the geodesic with initial velocity $v$. Since $d(\exp_p)_0=1$, the <inverse function theorem> makes $\exp_p$ a diffeomorphism near zero; its inverse gives <normal coordinates>. A geodesic sphere is $\exp_p(\{v:|v|=r\})$ inside such a normal neighborhood.
The <Gauss lemma> states
$$
\langle d(\exp_p)_v(v),d(\exp_p)_v(w)\rangle=\langle v,w\rangle.
$$
For the variation $\gamma_s(t)=\exp_p(t(v+sw))$, let $J=\partial_s\gamma_s|_{s=0}$. The coordinate vector fields commute, so metric compatibility and constant geodesic speed give
$$
\frac d{dt}\langle\dot\gamma,J\rangle
=\frac12\partial_s|\dot\gamma_s|^2\big|_{s=0}
=\langle v,w\rangle.
$$
Since $J(0)=0$, evaluation at $t=1$ proves the formula. In particular radial and spherical directions are orthogonal.
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