Let $(\mathcal{M}, g)$ be a complete manifold. Let $w\in \mathcal M$, $\xi_w\in T_w\mathcal M$ and $$v= \operatorname{Exp}_w (\xi_w),$$ where $\operatorname{Exp}(\cdot)$ is the exponential map. Now let $$w' = \text{Exp}_v (\xi'_v),$$ where $\xi'_v \in T_v\mathcal{M}$.
Assuming the inverse exponential map $\text{Exp}^{-1}$ is well defined, can we say that $\text{Exp}_w^{-1}(w') = \xi_w + T_v^w\xi'_v$, where $T_v^w$ is the parallel transport?
This is true in the simplest case, when $\mathcal M = \mathbb R^N$ is the Euclidean space with Euclidean metric. In this case, $\operatorname{Exp}_w(X) = w+X$ and $T^w_v = I$. Thus $$v = w + \xi_w, w' = v + \xi_v,$$ so $$w' = w + \xi_w + \xi_v \Rightarrow\operatorname{Exp}_w^{-1} (w') = \xi_w + \xi_v $$