How does the Sutherland-Hodgman Algorithm work?
As written in book:


  Let edge $E$, determined by endpoints $A$ and $B$, be any edge of the positively oriented, convex clipping polygon.


It means, that first of all you need orient your red rectangle counter-clockwise:



  We clip edge of the polygon in turn against the edge E of the clipping polygon, forming a new polygon. ... The algorithm proceeds in stages by passing each clipped polygon to the next edge of the windows and clipping


I.e.:


Clip green polygon by $\overline{WX}$ (polygon will stay the same);
Clip polygon from step 1 by $\overline{XY}$ (still the same polygon);
Clip polygon from step 2 by $\overline{YZ}$ (yellow part will be cut off, and result of this step will be polygon $ABCKLE$);
Clip polygon from step 3 by $\overline{ZW}$ (blue part will be cut off, and result of this step will be polygon $ABCKLMN$);


Here we can answer on your (1) question - yes, $\overline{AB}$ will be tested against all edges of clipping windows. And answer for question (4): $\overline{DE}$ will not be in resulting polygon, but $\overline{LM}$ (everything that lefts from $\overline{DE}$) will belong to result.

To address question (2) - we need to have a look at this quote (definition when point is on the left side or on the right side):


  Let $A(x_1,y_1)$ and $B(x_2,y_2)$ be the endpoints of a directed line segment. A point $P(x, y)$ will be to the left of line segment if the expression $C=(x_2-x_1)(y-y_1)-(y_2-y_1)(x-x_1)$ is positive.


$\overline{WX}$ is directed line segment and points $A$ and $B$ is on the left side, then whole $\overline{AB}$ edge is to the left of the line segment (cliping edge) $\overline{WX}$.

Question (3): again due to edge of clipping windows is directed $\overline{BC}$ really on the left side of $\overline{XY}$, $\overline{ZW}$ (also it's to the left of $\overline{YZ}$ and $\overline{WX}$).

I think, the cause of the questions that you forget to orient clipping rectangle.