I don't know if this would help but I'll try. In my mind, there is something we can do to compare Lie derivative and parralel transportation (that relies on covariant derivative).
Say $p \in M$ is a point in a riemannian manifold $M$ and that $\exp_p$ is its exponential map. Use polar coordinates for $T_pM$, whre $\mathbb{S}$ is the unit sphere in $T_pM$: this provides a map
\begin{align}
E : (0,A) \times \mathbb{S} & \longrightarrow M \\
(t,v) &\longmapsto \exp_p(tv)
\end{align}
which is a diffeomorphism on its image for small $A$ (because $\mathrm{d}\exp_p(0)$ is invertible). Think of that map as a chart for $M$. Its differential is
\begin{align}
\mathrm{d}E : T(0,A) \times T\mathbb{S} \to TM.
\end{align}
Choose $v$ a unitary vector in $\mathbb{S}$. Then $\gamma(t) = E(t,v)$ is a unitary geodesic. In this chart, it is just the function $t \mapsto (t,v)$, and the vector field $(1,0) \in T_t(0;A)\times T_v\mathbb{S}$ is sent by $E$ on $\gamma'(t)\in T_{\gamma(t)}M$. You can, by this chart, identifiate the derivative on the left hand side with respect to the coordinate $t$ with the Lie derivative on the right hand side with respect to $\gamma'(t)$. Moreover, the Gauss lemma says that $\{0\}\times T_v\mathbb{S}$ is identified with $\gamma'(t)^{\perp}$.
We have found a good frame to look at $TM$ : take any orthonormal frame of $v^{\perp} \in T_pM$ and transport it whith constant coordinates in this chart. It gives a frame along $\gamma(t)$ that is "constant" with respect to the Lie derivative $\mathcal{L}_{\gamma'(t)}$.
Thus, what is exactly a vector field along $\gamma$ which is constant in this chart? You can show they are Jacobi fields: knowing a bit of the curvature of $M$ and you can know many things about this chart.
More precisely: of $w \in T_v\mathbb{S}$, then $t \mapsto \mathrm{d}E(t,v)tw$ is the normal Jacobi field along $\gamma$ with initial data $Y(0) = 0$ and $\nabla_v Y(0) = w$. Thus, $\mathrm{d}E(t,v) w = \dfrac{Y(t)}{t}$ for the only solution of
\begin{align}
\nabla_{\gamma'(t)}\nabla_{\gamma'(t)}Y(t) +R\left(\gamma'(t),Y(t)\right)\gamma'(t)=0
\end{align}
with the desired initial data, where $R$ stands for the riemannian curvature tensor (careful : the sign depends on the convention. I choose $R(u,v) = \nabla_{[u,v]} - \left[\nabla_u,\nabla_v\right]$).
On the same time, you can create another frame along $\gamma(t)$ wich is just the parallel transport along this curve. What does this frame looks like in the chart $E$?
I think one can understand your question by "can we compare these two frames ?" or "what are the coordinates of the parallel frame in the chart $E$", but I may be completely wrong.
Attempt of application:
For a sphere of radius $R_0$, say $\mathbb{S}^{n+1}(R_0)$, choose a point $p$ and look at the exponential map. It provides a chart
\begin{align}
E : ]0,R_0\pi[ \times \mathbb{S}^{n} & \longrightarrow \mathbb{S}^{n+1} \\
(t, v) & \longmapsto \exp_p(tv)
\end{align}
(By the way: when you look at the metrics in these coordinates, you have $ E^*g = \mathrm{d}t^2 + {(R_0)}^2\sin^2\left(\dfrac{t}{R_0}\right) g_{\mathbb{S}^n}$ where $g_{\mathbb{S}^n}$ is the standard metrics on the $n$ dimensional sphere. It is pretty much what you wrote on your post!)
Take $w \in T_v\mathbb{S}^n$, and say you continue it by a parallel vector field $w(t)$ along $\gamma(t)=\exp_p(tv)$. Then it is easy to show that the Jacobi field with initial data $Y(0)=0$ and $\nabla_vY(0) = w$ is just $Y(t) = R_0\sin\left(\dfrac{t}{R_0}\right)w(t)$. It is because the curvature here is constant and equals $\frac{1}{{R_0}^2}$, thus the Jacobi equation is
\begin{align}
\nabla_{\gamma'(t)}\nabla_{\gamma'(t)} Y = -\dfrac{Y}{{R_0}^2}
\end{align}
and you can solve it in a parallel orthonormal frame along $\gamma$: in this frame, the covariant derivative corresponds exactly to the usual derivative on each component.
Thus the "Lie transportation" (in the sense stated above) of $w$ is $W(t) = \dfrac{\sin\left(\dfrac{t}{R_0}\right)}{\left(\dfrac{t}{R_0}\right)}w(t)$.
In that case, the Lie transportation is parallel to the parallel transportation but the curvature affects the norm!
Conversly, you can say that the parallel transport differs from the constant transportation by the multiplicative function $\dfrac{\left(\dfrac{t}{R_0}\right)}{\sin\left(\dfrac{t}{R_0}\right)}$.
My personal interpretation of this phenomena is that parallel transportation tries to be an isometry taking account of curvature. Thus the norm is respected. On the contrary, the exponential map tries to flatten your manifold, hence the constant vector fields in this map can look really hideous in the manifold! The real difference is that parallel transportation is a solution of a first order ODE which is $\nabla_{\gamma'}W=0$ while the Lie transportation is related to Jacobi fields, hence a second order linear ODE.
