The usual definition of the Fermi derivative (eg as given in Hawking and Ellis) is to consider a Lorentzian manifold $(M,g)$ and a unit timelike curve $\gamma$, a smooth vector field $X$ along $\gamma$, and define the Fermi derivative of $X$ along $\gamma$ to be the quantity \begin{align} \frac{D_{\text{Fermi}}X}{ds}&=\frac{DX}{ds}-g\left(X,\frac{D\dot{\gamma}}{ds}\right)\dot{\gamma}+ g\left(X,\dot{\gamma}\right)\frac{D\dot{\gamma}}{ds}, \end{align} where $\frac{D}{ds}$ denotes the covariant derivative along the path $\gamma$. Following this definition, Hawking and Ellis state that we can extend this to all tensor fields along $\gamma$ simply by requiring $\frac{D_{\text{Fermi}}}{ds}$ to commute with contractions, obey the Leibniz rule with respect to tensor products, and such that $\frac{D_{\text{Fermi}}f}{ds}=\frac{df}{ds}$ for smooth functions $f$.

My question is whether one can define an analogue of this in arbitrary vector bundles $(E,\pi,M)$, say equipped with a bundle metric $g$ (and say a corresponding metric-compatible connection on $E$)? I ask because for a general vector bundle, it is not possible to start with a curve $\gamma$ in the base manifold and lift it naturally to a curve in $E$. Next, I considered the following as a candidate definition:

Let $(E,\pi,M,g)$ be a vector bundle with a metric $g$ and a compatible connection. Let $\gamma:I\to M$ be a smooth curve and fix a lifting $\gamma_1:I\to E$, which is normalized (so $|g(\gamma_1,\gamma_1)|=1$ identically). For any smooth lifting $X:I\to E$ of $\gamma$ we define the Fermi derivative of $X$ with respect to $\gamma_1$ to be \begin{align} \frac{D_{\text{Fermi}}X}{ds}&=\frac{DX}{ds}-g\left(X,\frac{D\gamma_1}{ds}\right)\gamma_1+ g(X,\gamma_1)\frac{D\gamma_1}{ds}\tag{$*$} \end{align}

So, the curve $\gamma_1$ which I am fixing is playing the role of the velocity $\dot{\gamma}$. However, the problem I'm facing with this definition is that if I consider two vector bundles $(E_i,\pi_i,M,g_i)$ for $i=1,2$ and consider a single base curve $\gamma:I\to M$ and two fixed liftings $\gamma_i:I\to E_i$ and $X_i:I\to E_i$, then the Fermi derivative of $X_1\otimes X_2$ (a lifting of $\gamma$ to $E_1\otimes E_2$) with respect to $\gamma_1\otimes \gamma_2$ doesn't satisfy the Leibniz rule with respect to $\otimes$ anymore.

So, I'm wondering if there's a way to 'fix' the definition so that I have a general notion of Fermi-derivative on an arbitrary vector bundle from which I can prove that when specialized to the tangent bundle $TM$, and the corresponding tensor bundles, the Leibniz rule is obeyed. (I know that I can start with $(*)$ as a definition and then just force by hand for the Leibniz rule to hold, but I'm not a fan of such algebraic extensions, hence the question).

The usual definition of the Fermi derivative (eg as given in Hawking and Ellis) is to consider a Lorentzian manifold $(M,g)$ and a unit timelike curve $\gamma$, a smooth vector field $X$ along $\gamma$, and define the Fermi-Walker derivative of $X$ along $\gamma$ to be the quantity \begin{align} \frac{D_{\text{Fermi}}X}{ds}&=\frac{DX}{ds}-g\left(X,\frac{D\dot{\gamma}}{ds}\right)\dot{\gamma}+ g\left(X,\dot{\gamma}\right)\frac{D\dot{\gamma}}{ds}, \end{align} where $\frac{D}{ds}$ denotes the covariant derivative along the path $\gamma$. Following this definition, Hawking and Ellis state that we can extend this to all tensor fields along $\gamma$ simply by requiring $\frac{D_{\text{Fermi}}}{ds}$ to commute with contractions, obey the Leibniz rule with respect to tensor products, and such that $\frac{D_{\text{Fermi}}f}{ds}=\frac{df}{ds}$ for smooth functions $f$.

My question is whether one can define an analogue of this in arbitrary vector bundles $(E,\pi,M)$, say equipped with a bundle metric $g$ (and say a corresponding metric-compatible connection on $E$)? I ask because for a general vector bundle, it is not possible to start with a curve $\gamma$ in the base manifold and lift it naturally to a curve in $E$. Next, I considered the following as a candidate definition:

Let $(E,\pi,M,g)$ be a vector bundle with a metric $g$ and a compatible connection. Let $\gamma:I\to M$ be a smooth curve and fix a lifting $\gamma_1:I\to E$, which is normalized (so $|g(\gamma_1,\gamma_1)|=1$ identically). For any smooth lifting $X:I\to E$ of $\gamma$ we define the Fermi derivative of $X$ with respect to $\gamma_1$ to be \begin{align} \frac{D_{\text{Fermi}}X}{ds}&=\frac{DX}{ds}-g\left(X,\frac{D\gamma_1}{ds}\right)\gamma_1+ g(X,\gamma_1)\frac{D\gamma_1}{ds}\tag{$*$} \end{align}

So, the curve $\gamma_1$ which I am fixing is playing the role of the velocity $\dot{\gamma}$. However, the problem I'm facing with this definition is that if I consider two vector bundles $(E_i,\pi_i,M,g_i)$ for $i=1,2$ and consider a single base curve $\gamma:I\to M$ and two fixed liftings $\gamma_i:I\to E_i$ and $X_i:I\to E_i$, then the Fermi derivative of $X_1\otimes X_2$ (a lifting of $\gamma$ to $E_1\otimes E_2$) with respect to $\gamma_1\otimes \gamma_2$ doesn't satisfy the Leibniz rule with respect to $\otimes$ anymore.

So, I'm wondering if there's a way to 'fix' the definition so that I have a general notion of Fermi-derivative on an arbitrary vector bundle from which I can prove that when specialized to the tangent bundle $TM$, and the corresponding tensor bundles, the Leibniz rule is obeyed. (I know that I can start with $(*)$ as a definition and then just force by hand for the Leibniz rule to hold, but I'm not a fan of such algebraic extensions, hence the question).