I believe that (2) and (3) will often fail. For example, let $\mathbf{G} = \mathrm{SL}(3,\mathbb{H})$, where $\mathbb{H}$ is the algebra of quaternions. This is an almost-simple algebraic group over $\mathbb{R}$. Consider
$$\mathbf{G}_\alpha = \begin{bmatrix} * & * & 0 \\ * & * & 0 \\ 0 & 0 & 1 \end{bmatrix},
\ \mathbf{G}_\beta = \begin{bmatrix} 1 & 0 & 0 \\ 0 & * & * \\ 0 & * & * \end{bmatrix},
\ t_\omega = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1 \end{bmatrix}, $$
where $\omega$ is a unit quaternion. Then $t_\omega$ is in $\mathbf{G}_\alpha \cap \mathbf{G}_\beta$, so (2) fails. In fact, $t_\omega$ is in $\mathbf{Z}_\alpha \cap \mathbf{Z}_\beta$, since it centralizes the maximal $\mathbb{R}$-split torus consisting of real diagonal matrices, so (3) fails.

Here is a guess at the answer to (1) when $F$ has characteristic zero. (I do not have the expertise to speculate about fields of positive characteristic.)

Let $\mathbf{T}$ be a maximal torus that contains $\mathbf{S}$. It is easy to see that $\mathbf{U}_\alpha$ is normalized by $\mathbf{T}$, so the Lie algebra $\mathfrak{u}_\alpha$ is a sum of root spaces for $\mathbf{T}$. Let $\psi$ be the smallest quasi-closed set of roots that contains all of the roots that occur in either $\mathfrak{u}_\alpha$ or $\mathfrak{u}_{-\alpha}$. It seems to me that $\mathbf{G}_\alpha$ will usually, if not always, be the corresponding almost simple subgroup $\mathbf{G}^*_\psi$.

As a special case, $\alpha$ could be a circled root in the Tits-Satake diagram. Removing all of the other circled roots yields a diagram that may be disconnected, and I suspect that $\mathbf{G}_\alpha$ may be the almost-simple group corresponding to the connected component that contains $\alpha$.

At least, this works for $\mathrm{SL}(n,\mathbb{H})$. Take, for example, $n = 3$. The Tits-Satake diagram is
${\bullet}{-}{\circ}{-}{\bullet}{-}{\circ}{-}{\bullet}$.
Let $\alpha$ and $\beta$ be the two circled vertices. After deleting $\beta$ (the rightmost circled vertex), there are two connected components. One, which we will ignore, is an isolated black vertex. The other, which consists of the first three vertices (${\bullet}{-}{\circ}{-}{\bullet}$), and represents $\mathbf{G}_\alpha$, corresponds to the copy of $\mathrm{SL}(2,\mathbb{H})$ in the top left corner. This agrees with the above description of $\mathbf{G}_\alpha$.

Note that, in this example, $\mathbf{G}_\alpha$ is the subgroup corresponding to the first three vertices, and $\mathbf{G}_\beta$ corresponds to the last three vertices. Their overlap is the middle vertex, which corresponds to the set of elements $t_\omega$ mentioned above. In general, I think that if $\alpha$ and $\beta$ are adjacent to the same component of black vertices, then $\mathbf{G}_\alpha \cap \mathbf{G}_\beta$ will contain the anisotropic group corresponding to these black vertices.