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.
The $n$ lines that you remove can be seen as $2n$ half lines starting from the origin. Choose any $2n-1$ of those. For each one take a loop from a fixed basepoint which winds once around that half line (and "only" that). The homotopy classes of these loops generate freely the fundamental group of your space.
Let $a_1,\dots,a_{2n-1}$ be the generators described above.
The homotopy class of a loop that winds once around the half line that you didn't choose (and around "only" that) is $a_{\rho(1)}^{k_1}\cdots a_{\rho(2n-1)}^{k_{2n-1}}$ for some $k_i \in \{1,-1\}$ and some permutation $\rho$ of $\{1,\dots,2n-1\}$.
You can see this more easily if you first apply a deformation retract to $S^2$.
EDIT:
Let's put orientation and labels to the loops of your example ($\mathbb{R}^3$ with the $x$ and $y$ axes removed):
Then $d \simeq abc$.
Best Answer
Here is a geometric way to see this. To any ordered basis $(v_1,v_2,\ldots,v_n)$ of your vector space $V$ associate the "flag" of subspaces $V_0=\{0\}$, $V_1=\langle v_1\rangle$, $V_2=\langle v_1,v_2\rangle$, ... $V_n=\langle v_1,v_2,\ldots,v_n\rangle=V$. The Gram-Schmidt algorithm turns any such basis into an orthonormal basis $(b_1,\ldots,b_n)$ that gives rise to the same flag of subspaces. It is moreover the unique such basis (orthonomal and with the same flag) for which in addition each $b_i$, inside $V_i$, is on the same side of the hyperplane $V_{i-1}$ as the original basis vector $v_i$.
Now taking $V=\Bbb R^n$ we can identify $GL_n^+(\Bbb R)$ with the set of ordered bases $(v_1,v_2,\ldots,v_n)$ with $\det(v_1,v_2,\ldots,v_n)>0$, and $SO(n)$ with the set of ordered orthonormal bases $(b_1,b_2,\ldots,b_n)$ with $\det(b_1,b_2,\ldots,b_n)>0$. Now for such a basis $(v_1,v_2,\ldots,v_n)$ let $(b_1,\ldots,b_n)$ be the orthonormal basis associated to it under Gram-Schmidt, and simultaneously (or successively if you prefer) deform every $v_i$ linearly to $b_i$, as $t\mapsto (1-t)v_i+tb_i$. The intermediate vectors stay inside $V_i$, and since $b_i$ is on the same side as $v_i$, they never enter $V_{i-1}$. This means the deformed vectors stay linearly independent at all times, so the deformation takes place inside $GL_n(\Bbb R)$. As the determinant cannot vanish anywhere we have $\det(v_1,v_2,\ldots,v_n)>0\implies \det(b_1,b_2,\ldots,b_n)>0$ and we have a deformation retract of $GL_n^+(\Bbb R)$ to $SO(n)$. It is in fact a strong deformation retract: elements of $SO(n)$ remain fixed.