In my work I have a need for some kind of analogue of Taylor expansion of a vector field on Riemannian manifold $\mathcal{M}$. I came to such an expression: $$ F(\operatorname{exp}_p(v)) = \operatorname{d}\operatorname{exp}_p \vert_{v} \Big( F(p) + \nabla_{p}{F}v + \frac{1}{2}\nabla_{p}^{2}{F}(v, v) + o(||{v}||_p^2) \Big), $$ where $v \in B_r(0) \subset T_p\mathcal{M}$, $B_r(0)$ maps on $\mathcal{M}$ through $\operatorname{exp}_p$ diffeomorphically, $\nabla$ is the Levi-Civita connection, $\nabla_pF(v) = \nabla_vF$ for any vector $v$ from $T_p\mathcal{M}$. I haven't seen this formula anywhere in several papers/books on Riemannian geometry I've read and it made me think I've done some mistake while deriving this formula. Here is what I've done: consider function $$\tilde{F}(v) = \operatorname{d exp}^{-1}_{p} \vert_{\operatorname{exp}(v)} \circ F \circ \operatorname{exp}_p(v).$$ It is a function from $T_p\mathcal{M}$ to $T_p\mathcal{M}$, so we can use standart Taylor formula for expansion around $0$: $$\tilde{F}(v) = \tilde{F}(0) + \operatorname{d}\tilde{F} \vert_{0}(v) + \operatorname{d}^2\tilde{F} \vert_{0}(v, v) + o(||v||_p^2).$$ Now we'll show that $\operatorname{d}\tilde{F}\vert_{0} = \nabla_pF$. By the chain rule we have $$\operatorname{d}\tilde{F}\vert_{0} = \operatorname{d}\big(\operatorname{d exp}^{-1}_{p} \vert_{\operatorname{exp_p}(0)}\big)\vert_{F \circ \operatorname{exp_p}(0)} \circ \operatorname{d}F \vert_{\operatorname{exp_p}(0)} \circ \operatorname d \operatorname{exp_p} \vert_{0}.$$ Since $\operatorname{d} \operatorname{exp_p} \vert_{0} = \operatorname{Id}$ and since by invere function theorem $\operatorname{d exp}_p^{-1} \vert_{\exp_p(0)} = (\operatorname{d exp}_p \vert_{0})^{-1} = \operatorname{Id}^{-1} = \operatorname{Id}$, we have $$\operatorname{d}\tilde{F} \vert_{0} = \operatorname{d}F\vert_{exp_p(0)} = \operatorname{d}F \vert_{p}.$$ As I understand, $\operatorname{d}F \vert_p = \nabla_pF$ (this is the point of my doubts). Therefore, $\operatorname{d}^{i}F \vert_p = \nabla^{i}_pF$. So, $$\operatorname{d exp}^{-1}_{p} \vert_{\operatorname{exp}(v)} \circ F \circ \operatorname{exp}_p(v) = \tilde{F}(0) + \nabla_pF(v) + \nabla^2_pF(v, v) + o(||v||^2_p).$$ After using inverse function theorem and equality $\tilde{F}(0)=F(p)$ we get first expression. So I have following questions: is this formula wrong and if yes, where am I wrong and can someone tell me some source where analogous result is derived?
UPDATE:
I've found at least one mistake in above formulas (which may be the cause of absence of curvature tensor mentioned by Yuri in comments). It is the wrong use of the chain rule. Since $\operatorname{d exp}_p^{-1} \vert_{v}$ depends on $v$, differential of $\tilde{F}$ should look like this I guess (by the analogy with $\mathbb{R}^n$ case): $$\operatorname{d}\tilde{F}\vert_{0} = \operatorname{d}h\vert_{0} \circ F \circ \operatorname{exp_p(0)} + h(0) \circ \operatorname{d}F \vert_{\operatorname{exp_p}(0)} \circ \operatorname d \operatorname{exp_p} \vert_{0},$$ where h = $\operatorname{d exp}_p^{-1} \vert_{v}$. But this is only a guess, I cannot justify it using properties of $\nabla$. I know that covariant derivative obeys Leibniz's rule on tensor product, I also know that it commutes with contraction, but I can't see how it helps here. I thought it was better to open dedicated question for this: Chain rule with covariant derivative.