Since I'm lazy, I'm shamelessly referring to the following question:

Derivative of Exponential Map

I completely agree, as Michael Pauley answered, that $\phi'(t)$ is nothing but $J_t(1)$, where $J_t$ is the Jacobi field with $J_t(0)=\gamma'(t), \nabla_{\frac{\partial{x}}{\partial{u}}} {J_t}(0)=0$. But what is $\nabla_{\phi'(t)} {\phi'(t)}?$

Notice that, we can think of $\phi(t)=J_t(1)$ as a "parallel translate" of the geodesic $\gamma(t)$. Indeed in the Euclidean spaces or their quotient spaces, the answer is $\nabla_{\phi'(t)} {\phi'(t)}=0$, and the "parallel translate" is a geodesic, which is a staright line in Euclidean space. But what can we say if we work with general manifolds?

I've two questions:

1. Are there milder conditions on the manifold, than being Euclidean, that can guarantee that $\phi(t)$is a geodesic?

2. Are there computations that could relate $\nabla_{\phi'(t)} {\phi'(t)}$ to $\gamma, J, R$, $R$ being the curvature tensor, which I'm assuming will arise in the expression.

You may ignore the rest.

What I tried to do so far, without much success:

On P. 174, Lemma 10.1 of of John Lee's Riemannian manifolds and curvature book (available for free download online), I tried to put $V=S$ and used the symmetry lemma: $D_s{T}=D_t{S}$ to obtain:

$D_s^2 (T) - D_t(D_s(S)) = R(S,T)S$. Note that $D_s(S)=\nabla_{\phi'(t)} {\phi'(t)}$. But that's how far I'm only getting!

Thanks for your answers/inputs!