Choose $y \in L$ with $\sigma(y)=-y$.
Let's concretely look at the two examples case 1: $H = \pmatrix{0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0}$ and case 2: $H = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1}$.
Part 1: Seeing the standard Galois action $\alpha \mapsto \sigma \cdot \alpha$ on roots
Method 1: With the Lie algebra
Case 1: The Lie algebra of $G(k)$ is
$$\mathfrak g = \{ X \in M_4(L): Tr(X)=0 \text{ and } XH + H X^* =0\} = \{X \in M_4(L): Tr(X) = 0 \text{ and } x_{ij} = -\sigma(x_{5-j,5-i})\}$$
$$= \{ \pmatrix{a+by&c+dy&e+fy&gy\\h+iy&j-by&ky&-e+fy\\l+my&ny&-j-by&-c+dy\\oy&-l+my&-h+iy&-a+by} : a, ..., o \in k \}.$$
Surely its scalar extension $\mathfrak g_K$ is (isomorphic to) the split Lie algebra $\mathfrak{sl}_4(L) = \{X \in M_4(L): Tr(X)=0\}$, but let us look closely how Galois operates on that; namely, it does not just act on the entries of the matrix (when written as above), but rather as
$$\sigma_{\mathfrak g_L} := \sigma \otimes id_{\mathfrak g} \qquad \qquad (1)$$
on $\mathfrak g_L:= L \otimes_k \mathfrak g$. That is, e.g. the matrix
$$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0} \in \mathfrak g_L$$ "correctly viewed" as an element of the tensor product $L \otimes_k \mathfrak g$ is
$$=\frac12 \otimes \underbrace{\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0}}_{\in \mathfrak g} + \frac{1}{2y} \otimes \underbrace{\pmatrix{0&y&0&0\\0&0&0&0\\0&0&0&y\\0&0&0&0}}_{\in \mathfrak g}$$
so that $$\sigma(\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0})=
\frac12 \otimes \pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0} + \frac{1}{2\color{red}{\cdot(-y)}} \otimes \pmatrix{0&y&0&0\\0&0&0&0\\0&0&0&y\\0&0&0&0} = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&-1\\0&0&0&0} $$
from which one gets $\sigma (\pmatrix{0&x&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0}) = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&-\sigma(x)\\0&0&0&0}$ for all $x \in L$. Likewise, for example,
$$ \sigma(\underbrace{\pmatrix{1&0&0&0\\0&-1&0&0\\0&0&0&0\\0&0&0&0}}_{=:H_{\alpha_1}})= \underbrace{\pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&-1}}_{=:H_{\alpha_3}})$$
whereas $\pmatrix{0&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&0}) \in \mathfrak g$ and hence is actually fixed by $\sigma$.
Now in this case, indeed the diagonal matrices in $\mathfrak g$ form a maximal toral subalgebra (a.k.a. Cartan subalgebra) which contain a maximal split toral subalgebra (namely, the two-dimensional $\mathfrak s := \{ \pmatrix{a&0&0&0\\0&j&0&0\\0&0&-j&0\\0&0&0&-a} : a,j \in k \}$), so we can use the diagonal as above to see how Galois acts on the roots. Turns out that $\sigma$ maps the root space $\mathfrak g_{L, \alpha_1}$ onto the root space $\mathfrak g_{L, \alpha_3}$ (I call $\alpha_i = \eta_i - \eta_{i-1}$ the standard roots of the diagonal in $\mathfrak{sl}_4(L)$); and, accordingly, interchanges the coroots $H_{\alpha_1}$ and $H_{\alpha_3}$; in short, now there are many ways to define the action of Galois on the roots via $\sigma \cdot \alpha_1 := \alpha_3$ and $\sigma \cdot \alpha_2 =\alpha_2$. The actual technical formula to do it is
$$\sigma \cdot \alpha := \sigma \circ \alpha \circ \sigma_{\mathfrak g_L}^{-1}$$
where the first $\sigma$ is just the normal Galois on $L$, but the second (inner) $\sigma_{\mathfrak g_L}$ is defined via $(1)$ and in particular is not just Galois action on entries, so that this normal Galois action $\alpha \mapsto \sigma \cdot \alpha$ is not trivial.
Case 2: Now the Lie algebra of $G(k)$ is
$$\mathfrak g = \{ X \in M_4(L): Tr(X)=0 \text{ and } X = - X^* \} = \{X \in M_4(L): Tr(X) = 0 \text{ and } x_{ij} = -\sigma(x_{j,i})\}$$
$$= \{ \pmatrix{ay&b+cy&d+ey&f+gy\\-b+cy&hy&i+jy&k+ly\\-d+ey&-i+jy&my&n+oy\\-f+gy&-k+ly&-n+oy&(-a-h-m)y} : a, ..., o \in k \}.$$
Again its scalar extension $\mathfrak g_K$ is (isomorphic to) the split Lie algebra $\mathfrak{sl}_4(L)$, but this time Galois operates on that via
$$\sigma_{\mathfrak g_L} : X \mapsto -X^* (= -\sigma(X)^{tr}).$$
This seems to immediately give us an action $\sigma \cdot \alpha_i = -\alpha_i$ ($i=1,2,3$) on the roots, however we have to be very careful here. The problem is that although again the diagonal matrices in $\mathfrak g$ define a maximal toral subalgebra (CSA) $\mathfrak t$, and certainly $\mathfrak t_L$ is a split CSA in $\mathfrak g_L$, this $\mathfrak t$ does not necessarily contain a maximal split toral subalgebra of $\mathfrak g$ (e.g. if $k$ is a $p$-adic field, one can show there are non-zero ad-diagonalisable elements in $\mathfrak g$, but obviously none is in $\mathfrak t$): and we need that to set up the proper Galois action on the roots.
But if we assume that $k=\mathbb R$, $y=i$ (imaginary unit), then one can show that $\mathfrak g$ has no non-zero split toral subalgebra at all, and we are good to go, and indeed $\sigma$ operates on the entire root system as $-id$.
Method 2: With tori and algebraic geometry
(References: Galois action on character group and Galois action on the character group of a torus, but I find these a bit misleading because they focus on cases where the Galois action is just the one on points / matrix entries. Actually Definition of the unitary group via a cocycle on $\operatorname{GL}_n$ shows better what's going on here.)
Case 1: Our group has a torus with $T(L) = \{ \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} : x_i \in L, x_1 x_2 x_3 x_4 =1\}$ which is not split in $G(k)$, rather we get
$$T(k) = \{ \pmatrix{a&0&0&0\\0&b&0&0\\0&0&\sigma(b)^{-1}&0\\0&0&0&\sigma(a)^{-1}} : a,b \in L, ab \in k\}.$$
Accordingly, the Galois action on $T(L)$ is again not given by $\sigma$ acting on entries, but rather via
$$\sigma_{T(L)} : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \pmatrix{\sigma(x_4)^{-1}&0&0&0\\0&\sigma(x_3)^{-1}&0&0\\0&0&\sigma(x_2)^{-1}&0\\0&0&0&\sigma(x_1)^{-1}}.$$
Again we define the $\sigma$-action on a character $\chi : T(L) \rightarrow L^\times$ of the torus via
$$\sigma \cdot \chi := \sigma \circ \chi \circ \sigma_{T(L)}^{-1}$$
and again we can see that rather than being trivial, e.g. the character $\chi_1 : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \dfrac{x_1}{x_2}$ (i.e. the root $\alpha_1$ written multiplicatively) gets sent to
$$ \sigma \cdot \chi_1 : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \dfrac{x_3}{x_4}$$
i.e. the root $\alpha_3$ written multiplicatively, etc.
Case 2: (But again assuming that the maximal $k$-split torus of $G$ is trivial, e.g in the case $k=\mathbb R$, but not for $p$-adic fields $k$.) Again we can choose $T(L) = \{ \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} : x_i \in L, x_1 x_2 x_3 x_4 =1\}$ but this time in $G(k)$ we get
$$T(k) = \{ \pmatrix{a&0&0&0\\0&b&0&0\\0&0&c&0\\0&0&0&d} : a,b,c,d \in L, abcd =1, a^{-1} = \sigma(a), b^{-1}=\sigma(b), c^{-1}=\sigma(c)\}.$$
Accordingly, the Galois action on $T(L)$ is again not given by $\sigma$ acting on entries, but rather via
$$\sigma_{T(L)} : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \pmatrix{\sigma(x_1)^{-1}&0&0&0\\0&\sigma(x_2)^{-1}&0&0\\0&0&\sigma(x_3)^{-1}&0\\0&0&0&\sigma(x_4)^{-1}}.$$
Again we define the $\sigma$-action on a character $\chi : T(L) \rightarrow L^\times$ of the torus via
$$\sigma \cdot \chi := \sigma \circ \chi \circ \sigma_{T(L)}^{-1}$$
and again we can see that rather than being trivial, e.g. the character $\chi_1 : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \dfrac{x_1}{x_2}$ (i.e. the root $\alpha_1$ written multiplicatively) gets sent to
$$ \sigma \cdot \chi_1 : \pmatrix{x_1&0&0&0\\0&x_2&0&0\\0&0&x_3&0\\0&0&0&x_4} \mapsto \dfrac{x_2}{x_1}$$
i.e. the root $\color{red}{-}\alpha_1$ written multiplicatively, etc.
Part 2: The $*$-action
(References: 6.2 in Borel-Tits, 2.3 in Tits' Classification article in the Boulder Proceedings, ...)
Now this can be done in various ways, using maximal parabolic subgroups and whatnot, but here is a way of doing it just on the root level:
Say we are given a root system $R$ in an Euclidean space $V$, and an action of a finite Galois group $\Gamma$ via automorphisms of $R$.
Set $V_0 := \{v \in V: \sum_{\sigma \in \Gamma} \sigma \cdot v =0 \}$. (To make things work, here it is crucial that in the above one had chosen a maximal Torus $T$ which contains a maximal $k$-split torus $S$: Because then and only then will $V/V_0$ be well-behaved enough and realise the $k$-relative roots as quotient ("folding") of the original root system $R$.)
A $\Gamma$-linear order on $R$ (or $V$) is a choice of a positive half-space in $V$ such that for all $v \in V \setminus V_0$, we have $v$ positive $\implies \sigma \cdot v$ positive for all $\sigma \in \Gamma$. One can show that such $\Gamma$-linear orders always exist.
A $\Gamma$-basis for our root system $R$ is a set of simple roots among the ones positive for a $\Gamma$-linear order. I.e. a basis consisting of roots which are positive for a $\Gamma$-linear order.
Examples: In case 1, there are eight possible $\Gamma$-bases: $\{\alpha_1, \alpha_2, \alpha_3\}$ and its negative, $\{\alpha_1, -\alpha_1-\alpha_2-\alpha_3, \alpha_3\}$ and its negative, $\{\alpha_1+\alpha_2, -\alpha_2, \alpha_2+\alpha_3\}$ and its negative, and a fourth one I leave to you. Whereas e.g. $\{\alpha_2, \alpha_3, -\alpha_1-\alpha_2-\alpha_3\}$ is a basis, but not a $\Gamma$-basis of $R$.
In case 2, every basis of the root system is a $\Gamma$-basis (because $V=V_0$, the extra condition for positive systems to be $\Gamma$-linear is empty).
Now the $*$-action is defined as follows: For any basis $\Delta$ of the root system, $\sigma \cdot \Delta$ is again a basis. As is well-known, the Weyl group operates simply transitively on the set of bases of the root system; further, let's take a $\Gamma$-basis for $\Delta$. Then there is a Weyl group element $w_{\sigma}$ (actually, a unique one) such that $w_{\sigma}(\sigma \cdot \Delta) = \Delta$. Then we define the $*$-action as
$$\sigma * \alpha = w_{\sigma} (\sigma \cdot \alpha).$$
So to say, the $*$-action is the normal Galois action but "bent back" from a $\Gamma$-basis into itself via the Weyl group.
Now in our examples, in case 1, the normal action already stabilises each $\Gamma$-basis, and so the $*$-action is just the normal action. In case 2, the normal action actually sends each basis to its own negative, so as $w_\sigma$ we have to take the longest element of the Weyl group, which again gives the non-trivial automorphism of the Dynkin diagram.
In both cases, the $*$-action of the non-trivial element of the Galois group is given by $\alpha_1 \mapsto \alpha_3, \alpha_2 \mapsto \alpha_2$.
Now you will notice that a lot of this seems like overkill. Indeed, in both cases, it seems as if one does not need $\Gamma$-bases at all! Indeed, one can readily tell whether the "normal" Galois action acts through the Weyl group or not, and if not, then it necessarily gives the non-trivial diagram automorphism.
I am sure though that in more complicated examples (different root systems but more importantly, more intricate Galois groups!) this will make a difference. And then it is a crucial fact remarked in the above sources that the $*$-action, unlike the normal action, does not depend on many choices, in particular of the involved tori.
