This is explained by the Borel-Tits relative structure theory for connected reductive groups over arbitrary fields. In particular, there is no need whatsoever to assume $F$ is perfect (recall that all global fields and all non-archimedean local fields should be treated on equal footing whenever possible, and the global and local function fields are never perfect). The explanation seemsbelow is long when written out, but the underlying principles are rather natural and clean.
Moreover, we will find a collection of root vectors satisfying the stronger requested Galois-equivariance property as near the start of the question. I will use Roman letters to denote roots, so "$a$" rather than "$\alpha$". I should begin by correcting what seems
Consider a relative root $a$ in a chosen basis for the relative root system (it corresponds to be a misconception:Galois orbit in a basis of the correct definitionabsolute root system such that the members of the orbit have nontrivial restriction to a maximal split torus, as we review below). The $a$-root space may have massive dimension as an $F$-vector space, but we will show that it is naturally a line over a finite separable extension $F_a$ showsof $F$. The construction of this enhanced linear structure over such an extension will make clear the sense in which $F_a/F$ and that $F_a$-linear structure refining the given $F$-linear structure is canonical for non-multipliable $a$ (and nearly canonical
for multipliable $a$). This will use in a crucial way the quasi-split hypothesis.
First, we fix what seems to be a misconception: the correct definition of $F_a$ will show that it is generally not a Galois extension of $F$. Rather, it is a finite separable extension of $F$ (canonical when $a$ is non-multipliable in the relative root system, but seemingly only quadratic Galois over something canonical in the multipliable case -- if anyone can see a way to canonify these latter cases then please let me know!), and the Galois orbit of absolute roots in question is not at allnaturally indexed by the Galois group for $F_a$ over $F$ (which won't make sensethere is no natural base point for the orbit, and no such Galois group when $F_a/F$ is not Galois) but rather is naturally indexed by the set of $F$-embeddings of
$F_a$ into a fixed separable closure $F_s$ (sanity check: ${\rm{Gal}}(F_s/F)$ does
act naturally and transitively on that set, regardless of whether or not $F_a/F$ is Galois!).
We are going to postpone the quasi-split hypothesis for as long as possible, to identify most clearly where exactly this assumption is truly essential. But at the risk of ruining all suspense, it should be said
that we will prove that for a relative root $a$ in a chosen basis for the relative root system (these correspond exactly to a Galois orbit in a basis of the absolute root system consisting of absolute roots with nontrivial restriction to a maximal split torus), the $a$-root space that may have massive dimension as an $F$-vector space is a line over a finite separable extension $F_a$ of $F$. The construction of this enhanced linear structure over such an extension will make clear the sense in which it is canonical for non-multipliable $a$ (and nearly so
for multipliable $a$), and will use in a crucial way the quasi-split hypothesis.
ForWe are going to postpone the quasi-split hypothesis for as long as possible, to identify most clearly where exactly this assumption is truly essential. For an arbitrary connected reductive $F$-group $G$ (not yet assumed to be quasi-split, to clarify ideas), let $S$ be a maximal split $F$-torus. The Borel-Tits relative structure theory over arbitrary fields ensures that:
Proposition. If $H$ is a connected semisimple group over a field $k$ and it is absolutely simpleabsolutely simple, simply connected, and quasi-split with $k$-rank equal to $1$ then $H \simeq {\rm{SL}}_2$$H$ is either $k$-isomorphic to ${\rm{SL}}_2$ or $H \simeq {\rm{SU}}_3(K/k)$$H$ is the quasi-split special unitary
group group ${\rm{SU}}_3(K/k)$ associated to a quadratic Galois extension $K/k$ that. In the latter case, $K/k$ is unique
updetermined by $G$ up to $k$-isomorphism.