This question pertains to Lemma 3.5 of [this article.][1] Let $M$ be a smooth Riemannian manifold and $\gamma$ some geodesic with respect to the Levi-Civita connection $\nabla$. For any $C^2$ vector field $Y$ along $\gamma$, define:

 - $x_Y(t) := \exp_{\gamma(t)}(Y(t))$ for all $t \in [0, T]$.
 - The $C^2$ homotopy $\tilde{x}_Y(h, t) := \exp_{\gamma(t)}(hY(t)).$
 - The Jacobi operator $H$. That is, $HX = \nabla_t^2  X + R_{\gamma}(X, \dot{\gamma})\dot{\gamma}$ for all $C^2$ vector fields along $\gamma$. Furthermore, it is clear that $\nabla_h \nabla_t \dot{\tilde{x}}_X \vert_{h = 0} = HX$
 - $\|Y\|_{2,\infty} = \displaystyle{\max_{t \in [0, T]}\{\|Y(t)\|, \|\nabla_t Y(t)\|, \| \nabla_t^2 Y(t)\|\}}.$

The Lemma in question is: 
For some $L > 0$ and for $t = 0$, if $Y$ is $C^{\infty}$ with $HY(0) = 0$, then for all $h \in [0, 1]$, $$\|\nabla_t \dot{\tilde{x}}_Y(h, 0)\| \le L \|Y\|_{2, \infty}.$$

The proof goes as follows:

 1. $\dot{\tilde{x}}_Y(t) = E_1(hY, h\nabla_t Y, t) := d\exp_{x_{hY}(t)}(\dot{\gamma}, h\dot{Y})$, where $E_1$ is $C^{\infty}$ with all derivatives bounded. Similarly,
 2. $\nabla_t \dot{\tilde{x}}_Y(t) = E_2(hY, h\nabla_t Y, h \nabla_t^2 Y, t)$, where $E_2$ is $C^{\infty}$ with all derivatives bounded.


Then, for some $L > 0$ and for all $(h, t) \in [0, 1] \times [0, T]$, 
$$\frac12 \| \nabla_h^2 \nabla_t \dot{\tilde{x}}_Y(h, t)\| \le L \|Y\|_{2, \infty}.$$

The result now follows from Taylor's theorem, $\nabla_h \nabla_t \dot{\tilde{x}}_Y \vert_{h = 0} = HY$, and the fact that $HY(0) = 0$.

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This argument doesn't make much sense to me, I seem to be missing a lot of details. Why are all the derivatives of $E_1$ and $E_2$ necessarily bounded? How do $1$ and $2$ imply the bound on $\nabla_h^2 \nabla_t \dot{\tilde{x}}_Y(h, t)$? What even is Taylor's theorem in this context and why does it imply the result?

At least naively, I could see writing something to the affect of 
\begin{align*}
\nabla_t \dot{\tilde{x}}_Y(h, 0) &= \nabla_t \dot{\tilde{x}}_Y(0, 0) + \nabla_h \nabla_t \dot{\tilde{x}}_Y(0, 0)h + \frac12 \nabla_h^2 \nabla_t \dot{\tilde{x}}_Y(0, 0) h^2 + O(h^3) \\
&= \nabla_t \dot{\gamma}(0) + HY(0) + \frac12 \nabla_h^2 \nabla_t \dot{\tilde{x}}_Y(0, 0) h^2 + O(h^3) \\
&= \frac12 \nabla_h^2 \nabla_t \dot{\tilde{x}}_Y(0, 0) h^2 + O(h^3)
\end{align*}
since $\gamma$ is a geodesic and $HY(0) = 0$. The last term is bounded as we want, but I have no idea how the order $h^3$ terms will behave, and this still doesn't explain how the bound on $\frac12 \|\nabla_h^2 \nabla_t \dot{\tilde{x}}_Y(h, 0)\|$ was obtained nor am I sure that this kind of expansion holds true.


  [1]: https://www.intlpress.com/site/pub/files/_fulltext/journals/cms/2014/0012/0008/CMS-2014-0012-0008-a002.pdf