The definition of half-edges in an article puzzles me I read the following article and was confused by the definition of half-edges given by the authors.

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*J. Pach, G. Tóth,  Graphs drawn with few crossings per edge. Combinatorica 17, 427–439 (1997). https://doi.org/10.1007/BF01215922
I have to introduce some definitions from the article for the reader's understanding (It may take some patience).
In a drawing of a multigraph, any two non-disjoint edges either share only
endpoints or have precisely one point in common, at which they properly cross.
Let $M$ be a multigraph drawn in the plane so that every edge crosses at most
$k$ other edges. Let $M'$ be a sub-multigraph of $M$ with the largest number of edges
such that in the drawing of $M'$ (inherited from the drawing of $M$), no two edges
cross each other. We say that $M'$ is a maximum plane sub-multigraph of $M$.
Remark: The deletion of all crossing edges of $M$ may produce isolated vertices, whether the isolated vertices are  included in $M'$, the author seems not to explain.
[half-edge] The closed portion between an endpoint of $e$ and the nearest crossing of $e$ with an edge of $M'$ is called a half-edge.

My understanding of a half-edge of a crossing edge is a part of crossing edge such that one end must be a vertex and the other end is a crossing. For example, as shown in the picture below, the two half-edges of the crossing edge $uv$ are $uc_1$ and $vc_3$. But I soon felt something was wrong. Because I can't understand the proof of Lemma 2.1 in the article.

Lemma 2.1. Let $0 \le k \le 4$ and let $M$ be a multigraph drawn in the plane so that
every edge crosses at most $k$ others. Let $M'$ be a maximum plane sub-multigraph
of $M$, and let $\Phi$ denote a face with  $|\Phi|\ge 3$ sides in $M'$, whose boundary is
connected. Then the number of half-edges in $\Phi$ is at most $(|\Phi|- 2)(k+ 1) - 1$.
Let  $ABC$ be a 3-face in the maximum plane sub-multigraph $M'$  and $k$ be equal to 1, if one vertex, says $A$, of the boundary of $ABC$ emanates a half-edge, the authors claim that the total number of half-edges in $ABC$ is $1$.
I can't understand why the total number of half-edges is $1$.  Why not $3$? The graphs below are what I tried to draw. Maybe $Ax$, $Bx$ and $Cx$ are all half-edges in $ABC$. So my understanding of half-edges seems to be wrong, what is the correct understanding of half-edges here?
For the second half of the definition of half-edges "with an edge of $M'$", I don't know what it means.

Edits:  I've just seen a similar link below, but the definitions of the two "half-edges" don't seem to be quite the same. (Maybe they are essentially the same, but I just don't see it.)

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*Definition of graph, using half-edges.

For example, a graph $M$ is drawn on the left of the figure below. So obviously, its maximum plane sub-multigraph $M'$ is obtained by removing all the crossing edges (all the red edges), as shown on the right side below.

Edit: Thanks nickgard for pointing out my misunderstanding of $M'$.

So what do all the half-edges in $M$ refer to?
 A: The definition in the article refers to a specific drawing in the plane. A half edge is a subset of the plane here. The definition on the other post is about something else.
I believe the problem with your drawing is that the green shaded area is in fact not a face, as the edge $BC$ crosses it.
A: 
My understanding of a half-edge of a crossing edge is a part of crossing edge such that one end must be a vertex and the other end is a crossing. For example, as shown in the picture below, the two half-edges of the crossing edge $uv$ are $uc_1$ and $vc_3$. But I soon felt something was wrong....


Almost. The other edge where a half-edge ends must be an edge in $M'$. Suppose the edges through $c_1$ and $c_2$ are in $M'$, but the edge through $c_3$ is not in $M'$ (just in $M$). Then the half-edges shown are $uc_1$ and $vc_2$, ignoring $c_3$. Taking $e=uv$, $c_1$ is a "crossing of $e$ with an edge of $M'$". $c_2$ is also. But $c_3$ is not a "crossing of $e$ with an edge of $M'$". So the half-edge containing $u$ goes to "the nearest crossing of $e$ with an edge of $M'$", which is $c_1$. The half-edge containing $v$ goes to "the nearest crossing of $e$ with an edge of $M'$", which is $c_2$.
I think this detail leads to the confusion in your $k=1, |\Phi|=3$ example.

I can't understand why the total number of half-edges is $1$.  Why not $3$? The graphs below are what I tried to draw. Maybe $Ax$, $Bx$ and $Cx$ are all half-edges in $ABC$.


The claim is about faces in $M'$. If $ABC$ is a face in $M'$, then edges $AB$, $AC$, and $BC$ are all in $M'$. But since $M'$ does not contain intersecting edges, this means edge $Av$ must not be in $M'$. $Av$ can still be in $M$, in which case the half-edge $Ax$ is in triangular face $ABC$. $Bx$ and $Cx$ are not half-edges since $Av$ is not in $M'$. More generally, edges in $M'$ never have half-edges. Every half-edge is part of the drawing of a removed edge, an element of $E(M) \setminus E(M')$.
For your example $M$ with six vertices:



That is one possible $M'$. $M'$ just needs to have the maximum number of edges possible with no intersections; or equivalently, the minimum number of edges possible are to be removed from $M$. You removed $(2,5)$ and $(3,6)$. We could also remove $\{(2,4),(2,5)\}$ or $\{(3,6),(4,6)\}$ to get a different "maximum plane sub-multigraph of $M$". (Obviously removing just one edge will never get a non-intersecting $M'$, since no one edge drawing goes through all three intersections.)
Sticking with your chosen $M'$, the removed edge $(2,5)$ has a half-edge between $2$ and $c_2$ and a half-edge between $5$ and $c_2$. The removed edge $(3,6)$ has a half-edge between $3$ and $c_3$ and a half-edge between $6$ and $c_3$.
For this $M$, $k=2$. If $\Phi$ is the triangular face $(2,4,6)$ of $M'$, then the claim is that the number of half-edges in $\Phi$ is less than or equal to
$$ (|\Phi|-2)(k+2) - 1 = (3-1)(1+2) - 1 = 2 $$
and that is in fact the number of half-edges in the triangle.
