This is our setup:
Let $(M,g)$ and $(\tilde{M},\tilde{g})$ be $n$-dimensional Riemannian manifolds. Pick two points $p\in M$ and $\tilde{p}\in\tilde{M}$, fix some linear isometry $$i:\mathrm{T}_p M\to\mathrm{T}_\tilde{p}\tilde{M},$$ let $V\subseteq M$ be a normal neighborhood of $p$ such that $\exp_\tilde{p}$ is defined on $i\circ\exp_p^{-1}(V)$, and define $$f:V\to\tilde{M},\ \ \ \ q\mapsto\exp_\tilde{p}\circ i\circ\exp_p^{-1}(q).$$ For all $q\in V$, there exists a unique normalized geodesic $\gamma:[0,t]\to M$ in $V$, with $\gamma(0)=p$ and $\gamma(t) = q$. Denote by $P_t$ the parallel transport along $\gamma$ from $\gamma(0)$ to $\gamma(t)$. Define $$\phi_t:\mathrm{T}_q M\to\mathrm{T}_{f(q)}\tilde{M},\ \ \ \ v\mapsto \tilde{P_t}\circ i\circ P_t^{-1}(v)$$ where $\tilde{P_t}$ is the parallel transport along the normalize geodesic $\tilde{\gamma}:[0,t]\to\tilde{M}$ given by $\tilde{\gamma}(0)=\tilde{p}$, $\tilde{\gamma}\,'(0)=i(\gamma'(0))$. Now, suppose we are given a Jacobi field $J$ on $\gamma$ such that $J(0)=0$, and assume (the conditions of the theorem in fact give us this) that $\tilde{J} = \phi_t\circ J$ is a Jacobi field along $\tilde{\gamma} = f\circ\gamma$.
Now, in the proof of this theorem (which is found in Do Carmo), we are told that $\tilde{J}\,'(0) = i(J'(0))$. But why is this so? It seems a bit bizarre for me to even see a derivative inside of another function – the opposite of what happens in the usual chain rule.
The chain rule should be applied to the relation $\tilde{J} = \phi_t\circ J$. The derivatives of the Jacobi fields are related by $d\phi_t$ (some authors use $T\phi_t$) and the identity you mentioned results from the definition of $i$ and the choice of $\tilde\gamma$.