Ok, let me answer these questions in the order you asked them. About almost flat, I could not understand the 'U' turns part of your question, but to see what Hatcher means by almost flat, you should think about this picture:
So, the only time our knot is not in the $x,y$ plane is at the crossing and the overarc just is bumped up (in the positive $z$ direction) slightly so it doesn't intersect the underarc. The arcs $\alpha_i$ are the pieces of the knot that are in the $x,y$ plane. In our picture, that is everything except the small bump in the red arc. The $\beta_l$'s are the bumps that come out of the plane. So, the whole knot is the union of all the $\alpha_i$'s and $\beta_l$'s.
Next, the square $S_l$ is not a subset of $R_i$. Here is a picture of the $R_i$'s before we add in $S_l$.
I would think of $S_l$ as continuing $R_j$ but it rides over $R_i$, so two edges of its boundary lies in $R_i$ but nothing else.
Last, now that we have $X$, we can think of "picking up $K$" as literally just shifting the knot up in the $z$ direction slightly, so that it is in the center of the space between the table top and $R_i$'s. So, in our picture, the knot is the red and blue arc, which used to lie in the green table, but we move it so it is in the center of the space we made with the $R_i$'s and $S_l$'s.
To imagine the retract, it is the same idea as taking $\mathbb{R}\setminus \{x $-axis$ \}$ and retracting to an infinite cylinder around the $x$-axis. For us, instead of the $x$-axis, we have a loop, so instead of a cylinder, we would get a torus, but the crossings make us identify parts of the torus. Also, we get the table top as part of our retract.
I hope this was helpful.
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