Expressing pushforward of a flow in integral form Let $\phi(t,x)$ be a flow of a vectorfield $V$ on some compact domain $\tilde{U} = U\times I \in R^n \times R$. Let X be a vector field. If one wants to write
$(\phi(t,x))_{*}X)(\phi(t,x)(q)) = \phi(t,x))_*(X(q))$
in integral form, is there anything wrong in writing
$(\phi(t,x))_*(X(q)) = X(q) + \int_{0}^t DV_{\phi(x,s)}\circ D\phi(x,s)X(q) ds$
(this can be obtained either by direct derivation of $\phi(t,x))_{*}(X(q))$ or by finding an ODE satisfied by the flow $(\phi(t,x))_{*}X)(\phi(t,x)(q))$ using chronological calculus)
so in particular, 
$|\phi(t,x))_*(X(q)) - X(q)| \sim t\sup|DV|\sup|D\phi||X|$
I am asking this because usually people dont express pushword in the integral form but they express it as a series
$\phi(t,x))_* = Id + t[V,.] + t^2[V,[V,.]]+...$
(which converges on a bounded subset of functions). Then continuing this analysis one can derive results where (see Agrachev - Control Theory from Geometric Viewpoint Chapter 2)
$|((\phi(t,x))_*-I)X| \sim e^{|DV|}|DX|$
The appearance of $|DX|$ in one analysis and $|X|$ in the other confuses me. Is there a problem in the first part? Thanks
 A: Based on the comment discussion, what you call $\phi_*(X(q))$ I usually think of as $\mathrm{d}\phi\cdot X(q) = X(\phi)(q)$. 
That is to say, writing $\phi$ as a $\mathbb{R}^n$ valued function, what you are interested in is the (spatial) directional derivative of $\phi(t,x)$ in the direction of $X$ evaluated at the point $q$. In coordinates where $X = X^j \partial_j$ it is just
$$ (X^j \partial_j \phi^i)(t,q) $$
The reason your formula doesn't agree with those in the textbooks is because your formula is looking at something different from what the textbooks are looking at. If you are estimating something entirely different from what the textbooks are estimating, why should you expect that you get the same result?
(In your case you are trying to compare two vectors one based at the point $q$ and the other at $\phi(t,q)$. In the general geometric setting this doesn't make too much sense, since the two vectors live in two distinct vector spaces and cannot be compared. In your Euclidean setting you can work in canonical coordinates and ignore the coordinate ambiguity.) 
In any case, what you need to estimate is the difference
$$ \partial_j\phi^i(t,q) - \delta_j^i $$
Using the partial derivatives commute, and that $\delta_j^i = \partial_j\phi^i(0,q)$ you have that
$$ \partial_j\phi^i(t,q) - \delta_j^i = \int_0^t \partial_t\partial_j\phi^i(s,q) \mathrm{d}s = \int_0^t \partial_j \left(V^i(s,\phi(s,q))\right) \mathrm{d}s $$
which is precisely what you derived. Using the chain rule we have
$$ = \int_0^t \partial_kV^i(s,\phi(s,q)) (\partial_j\phi^k(s,q) - \delta_j^k) + \partial_jV^i(s,\phi(s,q)) \mathrm{d}s $$
From this we can apply Gronwall and get
$$ |\partial_j\phi^i - \delta_j^i| \leq t \sup |DV| \exp (t \sup |DV|) $$
