Folding of curves in Tristan Needham's proof of Hopf Umlaufsatz 

Pg -178,
Let us illustrate and clarify these ideas with a concrete example. If C is a circle, then as x orbits
C once, its image $\tilde{x} = N(x)$ orbits $S^1$ once, with matching angular speed. Clearly, deg[N(C)] = +1.
Now suppose that we gradually and symmetrically deform the circle, so that C takes the form
shown on the left of [17.4], resembling the cross section of a pear.
For the illustrated point $\tilde{q}$ on S
1
there is precisely one preimage q on C. But for point $\tilde{p}$ there
are instead three preimages p1, p2, p3.
A helpful mental device in locating these preimages of $\tilde{p}$ is to first observe that the tangent
to S
1 at $\tilde{p}$(not shown) must be parallel to the tangent to C at each preimage of $\tilde{p}$. Now imagine
taking this tangent line at $\tilde{p}$ and letting it move parallel to itself towards C, ultimately sweeping
across all of C. Note each time the moving line touches C: these include all the preimages of $\tilde{p}$, but
they also include the preimages of the antipodal point −$\tilde{p}$
Restricting attention to the right hand side of C, we observe that there are precisely two inflection points, a and b, distinguished by the fact that C crosses the tangent line at these points, and
only at these points.
To see that the inflection points play a crucial role in relation to the spherical/normal map N,
imagine x starting at q and travelling up the right side of C. Then $\tilde{x}$ = N(x) travels forward along
S
1
from $\tilde{q}$ through be and $\tilde{p}$ until it hits $\tilde{a}$. But at $\tilde{a}$ it bounces and travels backwards, passing through
$\tilde{p}$ for a second time before hitting be. Now it bounces again and resumes its forward motion along
S
1
, passing through $\tilde{p}$ a third time. Next it arrives at $\tilde{a}$ for a second time
Pg-179, Now comes a crucial mental leap in the visualization of this motion: think of the motion of $\tilde{x}$ as the orbit of a bead travelling along a continuous, unbroken thread. Thus, as illustrated, when $\tilde{x}$ first arrives at $\tilde{a}$ and starts to move backwards on $\mathbb{S}^1$ , it can only be because the thread has folded back on itself a second time. Thus, as illustrated, as x passes through $p_1$, then $p_2$ , and finally $p_3$, $\tilde{x}$ passes through $\tilde{p}$ three times, first forward, then backward along the folded thread, then forward again along the twice-folded thread.
contin

I am having difficulty understanding what exactly is the meaning of "curve being folded back on itself" and "twice folded". From google, this is the definition of folding

bend (something flexible and relatively flat) over on itself so that one part of it covers another. source

It seems to me that the above definition is non sensical here because we are not covering anything with anything other. Maybe it could be that a different definition of folding is being used.
Could someone please explain what Tristan Needham is trying to convey in simpler words? Thanks.
 A: An illustration (using Geogebra) of a "curve being folded back on itself" or "twice folded" giving a "double cover".
Have a look at the following space curve, with equations:
$$x=\sin(t), \ \ y=\cos(t), \ \ z=\sin(t/2)$$
with a double point in $(1,0,0)$.
It is drawn on a cylinder with radius $1$.
Its period is $4 \pi$ ; it covers (in the sense of "spreading a vertical shadow") twice its projection, the unit circle, before coming back in the same position with the same tangent vector. This is a way to materialize the fact that a $2 \pi$ angle is not automatically the same as a $0$ angle...

A: I think I sort of understood it now. Here is an attempt at an explanation, suppose we parameterize the deformed circle by a parameter $t$ i.e: $\gamma:\left[0,1 \right] \to \gamma(t)$.
Let $\theta(t)$ be the function which measures the rotation of normal from a reference normal on the curve (you can choose any normal of a point on curve for this). To have folding simply means to have some collection points  with $\theta'(t)=0$.
Geometrically this translates to push out / pushing inside the circle as shown in the diagram.
