[step:Show that the fibrewise transports vary smoothly with $x$]
Define a bundle map $P: \iota_0^*(F^*E) \to \iota_1^*(F^*E)$ over $\operatorname{id}_X$ by
\begin{align*}
P(x,v) = (x,P_xv).
\end{align*}
We prove that $P$ is smooth locally on $X$.
Let $x_0 \in X$. For each $t \in I$, local triviality gives an open neighbourhood $W_t \subset X \times I$ of $(x_0,t)$ over which $F^*E$ is trivial. Since $\{x_0\} \times I$ is compact, choose $t_1,\dots,t_N \in I$ such that the corresponding sets $W_{t_j}$ cover $\{x_0\} \times I$. By the [tube lemma](/theorems/960) applied to the finite cover, after shrinking to an open neighbourhood $U \subset X$ of $x_0$ and choosing a subdivision
\begin{align*}
0 = \tau_0 < \tau_1 < \dots < \tau_M = 1,
\end{align*}
each closed subcylinder $U \times [\tau_{m-1},\tau_m]$ is contained in some trivializing neighbourhood $W_{t_j}$. It is therefore enough to prove smoothness of the transport map on one such subcylinder: the total transport from $0$ to $1$ is the finite composition of these local transport maps, with the transition maps between adjacent local frames inserted at the subdivision times, and each transition map is smooth. The endpoint values agree by uniqueness of parallel transport, so the finite composition is independent of the chosen local-frame description.
Fix one subinterval $J = [a,b] \subset I$ from this subdivision and a trivialization over $U \times J$. Let $r$ be the rank of $E$, and choose a smooth local frame
\begin{align*}
e_1,\dots,e_r: U \times J \to F^*E
\end{align*}
for $F^*E$ over $U \times J$. In this frame, a section along $\gamma_x$ has the form
\begin{align*}
s(t) = \sum_{a=1}^r u_a(t)e_a(x,t),
\end{align*}
where $u: J \to \mathbb{R}^r$ is the coefficient map defined by
\begin{align*}
u(t) = (u_1(t),\dots,u_r(t)).
\end{align*}
Define smooth coefficient functions $A_{ba}: U \times J \to \mathbb{R}$ by
\begin{align*}
\nabla_T e_a = \sum_{b=1}^r A_{ba}e_b.
\end{align*}
The parallel transport equation $\nabla_Ts=0$ becomes
\begin{align*}
\frac{d u_b}{dt}(t) + \sum_{a=1}^r A_{ba}(x,t)u_a(t)=0,
\qquad b=1,\dots,r.
\end{align*}
Equivalently,
\begin{align*}
\frac{d u}{dt}(t) = -A(x,t)u(t),
\end{align*}
where $A: U \times J \to \mathbb{R}^{r \times r}$ is the smooth matrix-valued map defined by
\begin{align*}
A(x,t) = (A_{ba}(x,t))_{b,a}.
\end{align*}
Let $I_r \in \mathbb{R}^{r \times r}$ denote the identity matrix. For the fixed subinterval $J=[a,b]$, let $\Phi_J: U \times J \to \mathbb{R}^{r \times r}$ be the fundamental matrix solution of the initial value problem
\begin{align*}
\frac{\partial \Phi_J}{\partial t}(x,t) = -A(x,t)\Phi_J(x,t).
\end{align*}
The initial condition is
\begin{align*}
\Phi_J(x,a) = I_r.
\end{align*}
By smooth dependence of ODE solutions on parameters, applied to this linear system with smooth coefficient matrix $A$, the map $\Phi_J$ is smooth in $(x,t)$. In the chosen frame, the transport from time $a$ to time $b$ is represented by the matrix $\Phi_J(x,b)$, so its local representative is the map $U \times \mathbb{R}^r \to U \times \mathbb{R}^r$ given by
\begin{align*}
(x,\xi) \mapsto (x,\Phi_J(x,b)\xi).
\end{align*}
This map is smooth. Composing the finitely many smooth local transports and smooth frame-transition maps from the subdivision gives a smooth representative for the full map $P$ near $x_0$. Since $x_0$ was arbitrary, $P$ is a smooth vector bundle map.
[/step]