I've been reading through Tits's Corvallis survey "Reductive groups over a local field" and something that surprised me that was taken for granted about the definition of root subgroups. Let $G$ be a reductive group over a local field $K$ with $K$-split maximal torus $S$. Given a relative root $\alpha \in \Phi(G,S)$, we of course have a root subgroup $U_\alpha$. However, none of the possible definitions of root subgroups I am aware of are adequate in general, in particular if both $\alpha$ and $2\alpha$ are relative roots. The thing that surprised me most initially is the implication that always $U_{2 \alpha} \subset U_\alpha$, which is something I don't see how to justify in general. So the main question I'm asking is, how is $U_\alpha$ defined?
To illustrate the issue, consider the simplest case possible: Let $G$ be the quasi-split form of $\mathrm{SU} _3$ for a quadratic, separable extension $L/K$. Then $G$ consists of the $L$-linear transformations of $L^3$ preserving the hermitian form $h(v_1, v_2, v_3) = v_1 \bar v_3 + v_2 \bar v_2 + v_3 \bar v_1$, where $\bar v$ is the Galois conjugate of $v \in L$. In terms of matrices, we have $g \in G$ if $g \in \mathrm{Res} _ {L/K} (\mathrm{SL} _3)$ and $g^\dagger J g = J$ where $g^\dagger$ is the transpose-Galois conjugate of $g$ and $$ J = \begin{bmatrix} 0&0&1\\ 0&1&0\\ 1&0&0 \end{bmatrix}. $$ One maximal $K$-split torus $S$ consists of the matrices of the form $$ \begin{bmatrix} a & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1/a \end{bmatrix}, \qquad a \in K. $$ Then $\Phi(G, S)$ is of type $BC_1$: there are $4$ relative roots of $(G, S)$, and the two positive roots differ by a scalar multiple of $2$. Denote the non-divisible positive root $\alpha$ the divisible one $2\alpha$, where $B$ is taken as the upper triangular matrices in $G$. In $\mathfrak g$, the Lie algebra of $G$, the root subalgebra $\mathfrak g_\alpha$ consists of matrices of the form $$ \begin{bmatrix} 0 & -\bar c & 0 \\ 0 & 0 & c \\ 0 & 0 & 0 \end{bmatrix}, \qquad c \in L, $$ and $\mathfrak g_{2\alpha}$ consists of matrices of the form $$ \begin{bmatrix} 0 & 0 & d \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \qquad d \in L \text{ such that } d + \bar d = 0. $$
There are 3 ways I'm aware of to define $U_\alpha$: one using the exponentials of the Lie algebra, one using root homomorphisms, and one using the semisimple subgroup $G_\alpha$ given as the derived subgroup of a certain Levi (but this last one is certainly not correct for non-reduced root systems). From what I've seen, the exponential definition gets closest, but I'm still not happy with it.
I would like to define $U_\alpha$ as $\exp{(\mathfrak g_\alpha)}$ in general for a relative root $\alpha$. This would always lead to $U_\alpha \cap U_{2\alpha} = \{1\}$, as $\mathfrak g_\alpha \cap \mathfrak g_{2\alpha} = 0$. However, working through the example above, one can see that $\exp{(\mathfrak g_\alpha)}$ is not a subgroup of $\mathrm{SU} _3$ for the non-divisible $\alpha$. So the next guess would be to define $U_\alpha$ as the subgroup of $G$ generated by $\exp{(\mathfrak g_\alpha)}$. If this is the correct definition, why is it always true that $U_{2\alpha} \subset U_\alpha$? Although that's true in this case, it seems totally plausible that for a non-divisible root $\alpha$, sometimes $\exp{(\mathfrak g_\alpha)}$ might be closed under group multiplication. It's also a bit unsatisfying because the exponential map is not defined in general in positive characteristic, and in particular if $\mathrm{char}(K) = 2$, then it's not defined for $\mathfrak g_{\alpha}$ above.
The next definition involves root homomorophisms. If one can find a define an injective homomorphism $u_\alpha: \mathbb G_a^n \to G$ such that $t u_\alpha(x) t^{-1} = u_\alpha(\alpha(t)\cdot x)$ for all $t \in S$ and $x \in \mathbb G_a^n$, then the image of $u_\alpha$ would seem to be a pretty good candidate for $U_\alpha$. However, in $\mathrm{SU} _3$, this is not possible for the non-divisible root $\alpha$. I think in general there's a scheme map from the additive group on $\mathfrak g_\alpha$ to $G$ that coincides with the exponential (or the closest thing possible given $K$), but since $\exp{(\mathfrak g_\alpha)}$ isn't necessarily a subgroup, this map is not a homomorphism. The root homomorphism Tits uses for $\mathrm{SU} _3$ is $$ u_1(c, d) = \begin{bmatrix} 1 & -\bar c & d \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}, \qquad c, d \in L \text{ such that } c\bar c + d + \bar d = 0, $$ which has the property that $t u_1(c, d) t^{-1} = u_1(\alpha(t) \cdot c, 2\alpha(t) \cdot d)$. This is clearly almost as good as the property I defined above and is consistent with $U_{2\alpha} \subset U_\alpha$. However, it seems to take this subset property for granted, and doesn't seem ideal for a definition of a root homomorphism for a non-divisible root—it seems more like a consequence of a different definition like exponentiation.
So what is the correct definition? Should I start with a root homomorphism or the Lie algebra, and either way, how can I make sure my definition always works?