$v$-adic expansions of non-$p$th powers in global fields

$\DeclareMathOperator\Der{Der}$Let $f\in \mathbb{F}_p[x]$. The assumption that $f$ is not a $p$-th power implies that the derivative $f'= df/dx$ is not identically zero. For all but finitely many places $v$, $f'$ will have valuation zero at $v$ as you write.

If $v$ corresponds to an irreducible polynomial $g\in \mathbb{F}_p[x]$ then $f'$ has valuation zero at $v$ if and only if $f'(\lambda)\neq 0$, where $\lambda$ is any zero of $g$ in $\overline{\mathbb{F}}_p$.

I think this implies the assertion, for if $\lambda$ is in the algebraic closure of $\mathbb{F}_p$ we have an equality in $\overline{\mathbb{F}}_p[[x-\lambda]]$:

$$f= f(\lambda) + f'(\lambda)(x-\lambda) + O((x-\lambda)^2).$$

To deduce from this that $a_{v,1}\neq 0$ with $\pi_v= g$ one has to use that $\mathbb{F}_p$ is perfect, so that every zero of $g$ occurs with multiplicity $1$ and hence $(dg/dx)(\lambda)\neq 0$, so that $a_{v,1}= f'(\lambda)/g'(\lambda)\neq 0$.

Added: I think the above proof works in general, provided we can find a substitute for the derivative $d/dx$. I think this can be done as follows.

Let $X$ be a smooth curve of finite type over a perfect field $k$ of characteristic $p$. The sheaf of Kähler differentials $\Omega_{X/k}$ is locally free of rank $1$. Let $U$ be an affine open of $X$, where $\Omega$ is free of rank $1$. Working on $U$ results in disregarding finitely many closed points on $X$ (=finitely many places of the function field).

Let $A$ be the ring of global sections of $U$. Then by construction $\Der_{k}(A, A)=\mathbb{Hom}_A(\Omega_X(U), A)$ is a free $A$-module of rank $1$. Let us choose a generator $d$ (=the analog of $d/dx$ in the previous proof).

If $\mathfrak p$ is a maximal ideal of $A$ then the image of $d$ generates $\Der_{k}(A_{\mathfrak p}, A_{\mathfrak p})$ as an $A_{\mathfrak p}$-module, and hence $d(\pi_v)\neq 0$, where $\pi_v$ is a uniformizer of $A_{\mathfrak p}$. In fact, we get that $d(\pi_v)$ is a unit in $A_{\mathfrak p}$.

From $f=\sum_{i\ge 0} a_{v,i} \pi_v^i$ we get

$$df =(\sum_{i\ge 1} a_{v, i} i \pi_v^{i-1}) d\pi_v.$$

Hence, $a_{v,1} = 0$ if and only if $df$ lies in the maximal ideal of $A_{\mathfrak p}$.

If $df \neq 0$ then by shrinking $U$ (which amounts to disregarding finitely many places) we may assume that $df$ has valuation $0$ at all $v\in U$. This allows us to conlude that if $df\neq 0$ then $a_{v,1}\neq 0$ for all $v\in U$.

It remains to make the condition $df=0$ explicit. It is equivalent to $f$ being a $p$-th power in every localization $A_{\mathfrak p}$. Is this equivalent to $f$ being a $p$-th power in $A$? I am not sure.

Expanding on my comment. (Edited to add: though, having seen Lubin's comment now, I agree that philosophically the right way to think about this is in terms of the differential form $df$ vanishing at only finitely many points on the algebraic curve.)

Consider the subfield $k' = \mathbb{F}(x)$ of $k$ (note that $x$ is a transcendental element of $k'$ because otherwise it would be a $p$th power).

I claim that the extension $k/k'$ is ramified at $v$ if and only if $a_{v,1} = 0$. There will be finitely many such places exactly when $k/k'$ is separable, which is equivalent to your condition that $x$ is not a $p$th power in $k$. (A proof of this fact can be found in Proposition 3.10.2 of Stichtenoth, Henning, Algebraic function fields and codes, Graduate Texts in Mathematics 254. Berlin: Springer (ISBN 978-3-540-76877-7/hbk). xiii, 355 p. (2009). ZBL1155.14022 where $x$ is called a "separating element" for $K$. Thanks to Arno Fehm's comment and his linked paper for helping me track down this reference!)

Without loss of generality we can assume that our base field $\mathbb{F}$ contains all coefficients $a_{v, i}$, since adjoining scalars to a field doesn't affect the ramification.

Now let $v$ be a place of $k$, and let $v'$ be its restriction to $k'$.

Rearranging your equation as

$$x - a_{v,0} = a_{v,1} \pi_v + a_{v,2} \pi_v^2 + \dotsb,$$

we see that $v(x - a_{v,0}) > 0$, and $v(x - a_{v,0}) > 1$ if and only if $a_{v,1} = 0$. The first statement implies that $v'$ is the place of $k' = \mathbb{F}_p(x)$ corresponding to the ideal $(x-a_{v,0}) \subset \mathbb{F}_p[x]$: in particular that means that $x - a_{v,0}$ is a uniformizer for the place $v'$.

We conclude that $v/v'$ is ramified if and only if $v(x - a_{v,0}) > 1$ if and only if $a_{v,1} = 0$.