Going with/against the orientation alternates along the Hamilton cycle?

In this answer, I assume that "correct" occurs when the HC path either arrives to a right-oriented node and turns right, or arrives to a left-oriented node and turns left. In other words, I assume that the behaviour of the HC at a given node is compared to the orienting of the same node. I specify this because the last example given in the OP could suggest that the behaviour of the HC at a given node is compared to the orienting of the successive node (however, I checked this alternative possibility and noted that it generates no regular alternance). In addition, I assume that "right-oriented" refers to a node where the $0/1/2$ color sequence of the relative edges is clockwise, and that "left-oriented" refers to a node where this sequence is counterclockwise.

Let us begin by considering the $i^{th} $ node of the HC, calling it $N_i $. Let us consider firstly the case in which the HC edge arriving to it from the preceding node $N_{i-1}$ has colour $1$ (and consequently the HC edge starting from $N_i $ towards $N_{i+1}$ has colour $2$).

Under these conditions, we can distinguish the following two subcases. If $N_{i}$ is a right-oriented node, then the HC turns on the left in $N_{i}$ (in fact, because a right-oriented node must have the $0/1/2$ colour edges in a clockwise order, it is necessary that the $0$-colour, non-HC edge goes on the right). On the other hand, if $N_{i}$ is a left-oriented node, then the HC turns on the right in $N_{i}$ (a left-oriented node must have the $0/1/2$ colour edges in a counterclockwise order, so that the $0$-colour, non-HC edge goes on the left). As a result, in both subcases the HC turns wrongly. This means that, along the HC, in any node that terminates a colour $1$ edge (and that starts a colour $2$ edge) the behaviour of the HC is wrong.

Now let us consider the inverse case, in which the HC edge arriving to $N_i $ has colour $2$, and that starting from $N_i $ has colour $1$. By considerations similar and specular to those above, we get that if $N_{i}$ is a right-oriented node, then the HC turns right in $N_{i}$, whereas if $N_{i}$ is a left-oriented node, then the HC turns left in $N_{i}$. As a result, in both subcases the HC turns correctly. This means that, along the HC, in any node that terminates a colour $2$ edge (and that begins a colour $1$ edge) the behaviour of the HC is correct.

Therefore, because the HC is by definition composed by edges with a regular alternance of colour $1$ and colour $2$, we get that its behaviour must necessarily follow a correct/wrong alternance as well.