= Solution
A smooth <vector field> is a smooth <section of a vector bundle> $X:M\to TM$ of the <tangent bundle>, so $X_p\in T_pM$. In a <manifold chart>, $X=X^i\partial_i$ with smooth coefficients. It acts on a <smooth function> by $Xf=X^i\partial_i f$. For a <diffeomorphism> $\alpha:M\to N$, the <pushforward of a vector field> is
$$
(\alpha_*X)_q=d\alpha_{\alpha^{-1}(q)}X_{\alpha^{-1}(q)}.
$$
The <Lie bracket of vector fields> is the commutator of their actions on <smooth functions>:
$$
[X,Y]f=X(Yf)-Y(Xf),\qquad [X,Y]^j=X^i\partial_iY^j-Y^i\partial_iX^j.
$$
The second derivatives of $f$ cancel, so this is again a <vector field>. The corresponding <Jacobi identity> is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.
A <local flow> of $X$ is a smooth map $(t,p)\mapsto\phi_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying $\phi_0(p)=p$ and $\partial_t\phi_t(p)=X_{\phi_t(p)}$. On a sufficiently small neighbourhood and time interval, each $\phi_t$ is a <diffeomorphism> onto its image, with inverse $\phi_{-t}$. Uniqueness of <integral curves of a vector field> gives $\phi_{s+t}=\phi_s\circ\phi_t$ wherever both sides are defined. These are local statements; no assumption of a <complete vector field> is required.
A smooth <tensor field> of type $(r,s)$ is a smooth section of $(TM)^{\otimes r}\otimes(T^*M)^{\otimes s}$, where the factors are the <tangent bundle> and <cotangent bundle>. The <flow definition of the Lie derivative of a tensor field> is
$$
\mathcal L_XT=\left.\frac d{dt}\right|_{t=0}\phi_t^*T.
$$
Here pullback applies $d\phi_t^{-1}$ to each vector factor and the dual of $d\phi_t$ to each covector factor, evaluating $T$ at $\phi_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.
For a <smooth function>, $\phi_t^*f=f\circ\phi_t$, and the <chain rule> gives
$$
\boxed{\mathcal L_Xf=df(X)=Xf}.
$$
For a <vector field> $Y$, in local coordinates the expansions $\phi_t(x)=x+tX(x)+O(t^2)$ and $d\phi_t(x)=I+t\,dX_x+O(t^2)$ give
$$
\phi_t^*Y(x)=d\phi_t(x)^{-1}Y(\phi_t(x))
=Y(x)+t\bigl(dY_xX(x)-dX_xY(x)\bigr)+O(t^2).
$$
Consequently
$$
\boxed{\mathcal L_XY=[X,Y]}.
$$
Pullback preserves <tensor products> and <tensor contractions>, so differentiation makes this definition a <tensor derivation>. It therefore agrees on every <tensor field> with the <Lie derivative of a tensor field> determined by these two formulas.
Now put $\psi_t=\alpha\circ\phi_t\circ\alpha^{-1}$. The <chain rule> gives
$$
\partial_t\psi_t(q)=d\alpha_{\phi_t(\alpha^{-1}q)}X_{\phi_t(\alpha^{-1}q)}
=(\alpha_*X)_{\psi_t(q)},\qquad \psi_0(q)=q.
$$
Thus $\psi_t$ is the <local flow> of $\alpha_*X$. In the case $M=N$, if $\alpha_*X=X$, uniqueness of <integral curves of a vector field> makes $\psi_t=\phi_t$ on their common domains, or $\alpha\circ\phi_t=\phi_t\circ\alpha$. Conversely, differentiating this commuting identity at zero gives $d\alpha_pX_p=X_{\alpha(p)}$, hence $\alpha_*X=X$. This proves that <diffeomorphism invariance of a vector field is equivalent to commuting with its local flow>, with every identity understood on the domain where its compositions exist.
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