Chain rule with covariant derivative

Question

Let $\mathcal{M}$ be a $n$-dimensional Riemannian manifold with Levi-Civita connection $\nabla$. Consider the following function: $$\tilde{F}(v) = \operatorname{d exp}^{-1}_{p} \vert_{\operatorname{exp}(v)} \circ F \circ \operatorname{exp}_p(v),$$ where $F$ is a vector field, $v \in B_r(0) \subset T_p\mathcal{M}$, $B_r(0)$ maps on $\mathcal{M}$ through $\operatorname{exp}_p$ diffeomorphicaly. I want to find its differential at $v=0$. Since $$\tilde{F}(v) = A(v) \circ B(v)$$ where $A$ and $B$ in coordinates are just matrix depending on $v$ and vector depending on $v$ respectfully, $\operatorname{d} \tilde{F}$ should obey product rule: $$\operatorname{d}\tilde{F}\vert_{v=0} = \operatorname{d}(\operatorname{d}\exp_p\vert_{(\cdot)})\vert_{v=0} \circ (F \circ \operatorname{exp}_p)(0) + \operatorname{d}\exp_p\vert_{0} \circ \operatorname{d}(F \circ \operatorname{exp}_p)\vert_{v=0},$$ First question: is it true that product rule can be used here and if yes, what exactly is $\operatorname{d}(F \circ \operatorname{exp}_p)\vert_{v=0}$? Is it by chain rule equals $\nabla_{\operatorname{exp}_p(0)}F \circ \operatorname{d exp}_p\vert_{0}$? By $(\nabla_p F)(v)$ I mean $(\nabla_vF)(p)$.