Understanding how coordinates change under the flow of a vector field

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Main Question or Discussion Point

[Ref. 'Core Concepts in Special and General Relativity' by Luscombe]

Let ##M,M'## be manifolds and ##\psi:M\to M'## a diffeomorphism. Even if ##\psi## weren't a diffeomorphism, and instead just a smooth map, the coordinates of the pushback of ##\mathbf{t}\in T_p(M)##, would be related to the coordinates of ##\mathbf{t}## by ##(\psi_*\mathbf{t})^a=t^i\frac{\partial y^a}{\partial x^i}## (more precisely it's ##y^a\circ\psi## but we ignore the composition in the usual notation.

Note that ##x^i## are the coordinates for the coordinate system covering ##p##, while ##y^a## are those for the coordinate system covering ##\psi(p)##. We thus get the relation $$\psi_*\mathbf{t}=t^i\frac{\partial y^a}{\partial x^i}\mathbf{e}_a$$
I'm dividing this into 3 parts for your convenience.

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Part 1

Now let a vector field ##\mathbf{v}## be given on ##M##. Let ##\phi_t:M\to M## be a diffeomorphism such that ##\phi_t(\gamma(r))=\gamma(r+t)##, where ##\gamma## is the integral curve associated with ##\mathbf{v}## that passes through ##\gamma(r)##. Then, quoting:
The Lie derivative of various objects is found by working out their pullbacks. Let ##p\in M## have coordinates ##x^{\mu}##. The coordinates of ##\phi_t(p)## are, for infinitesimal ##t##, generated by the vector ##\mathbf{v}## at ##p##, $$\phi_t(x^{\mu})\equiv \bar x^{\mu}=x^{\mu}+tv^{\mu}+O(t^2)$$
It hasn't been specified how the above came about, but I'm guessing it's because the manifold is locally flat in the infinitesimal region, and so we can related coordinates of nearby points as we usually do in vector calculus via directional derivatives in flat space. Is that correct?

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Part 2

So the above gives the coordinates of ##p'=\phi_t(p)##. Then it's written:
The pullback of a function at ##\bar x^{\mu}## (which I'm interpreting to mean that ##f## is defined as a smooth function in the neighborhood of ##p'##) is $$(\phi^*_t f)(x^{\mu})=f(x^{\mu}+tv^{\mu})\approx f(x^{\mu})+tv^{\mu}\partial_{\mu}f$$
My interpretation of the above is that ##(\phi^*_tf)(p)=f(\phi_t(p))=f(p')## and we want to find the coordinates of ##f(p')## in terms of those of ##p## and those of the vector ##\mathbf{v}|_p##. How exactly did we conclude that the components of ##f(p')## are given by the RHS? (Again my guess is that we're using the fact that the patch containing ##p## and ##p'## are locally flat, and some general multivariable form of Taylor expansion, but I'd still appreciate details this)

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Part 3

Let ##\phi_{-t}## be the inverse of ##\phi_t##. Now we come to evaluating the pullback (under ##\phi_t##) of a tangent vector at ##p'##. This is the same as the pushforward under ##\phi_{-t}##. Quoting:
The pushforward of a contravariant vector field ##\mathbf{u}## under ##\phi_{-t}## is, using Eq. (13.7) (the equation given at the start of this question), $$\phi_{-t\ *}(\mathbf{u}|_{\phi_t(p)})=u^a(x^{\lambda}+tv^{\lambda})\frac{\partial x^{\mu}}{\partial\bar x^a}\mathbf{e}_{\mu}$$
Do I interpret this to mean that ##v^a## represents the coordinates of the vector field ##\mathbf{v}## and not the vector ##\mathbf{v}|_{\phi_t(p)}##? Are the components of ##\mathbf{v}## to be considered functions of the coordinates of the point at which we're calculating the vector field?
 

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