I will try to answer your questions one by one with the help of certain diagrams that I have made. Before I begin, I would like to tell you that I am not an expert on this subject. In fact, I have just started studying this. So, any suggestions or methods other than the one I am going to explain here are, indeed, welcome.
Since your question is mostly about the second deformation retraction, we will concentrate on that. First, we see what we want to do. We want to map every point of a disc with two holes to a point in $\mathbb{S}^1 \cup \left( \left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right] \right)$. For the sake of simplicity, we will denote by $B_1 = \left\lbrace \textbf{x} \in \mathbb{R}^2 | \| \textbf{x} \| \leq 1 \right\rbrace \setminus \left\lbrace \left( \frac{1}{2}, 0 \right), \left( - \frac{1}{2}, 0 \right) \right\rbrace$. This is the unit disc with two points removed. Notice that you can do the same analysis that I will be doing by considering a disc of radius $2$ with two points removed (as you have mentioned in the beginning of your question).
As shown in the figure, we want to construct a function which brings the blue shaded region of $\mathbb{R}^2$ to the green shared region.
In the figure, the two red dots denote the removed points. In our case they are $\left( \frac{1}{2}, 0 \right)$ and $\left( -\frac{1}{2}, 0 \right)$. To understand how to construct a function as desired, let us first concentrate on the blue region alone.
From the figures we have been provided as intuitions, it is clear that we want to create a function using straight lines. That is, if there is a point $\left( x, y \right)$ in the right half-plane and inside $B_1$, then we want to look at the line joining $\left( x, y \right)$ and $\left( \frac{1}{2}, 0 \right)$. Then, we would want to map this point $\left( x, y \right)$ to the intersection of the line with the theta space.
This intersection can happen in two cases: One where the line intersects with $\mathbb{S}^1$ and the other, when it intersects with the line segment $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$. We want the intersection that happens first. That is, we would like to study two classes of lines. I have denoted them in the figure by orange and purple colours. The orange lines are those which intersection the line segment $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$ and the purple lines are those which intersection with the circle $\mathbb{S}^1$.
Let us look at the orange lines first. Of all the possible orange lines, two of them intersect the theta space at $\left( 0, 1 \right)$ and two intersect the theta space at $\left( 0, -1 \right)$. These four line segments make a quadrilateral. Anything inside this quadrilateral will have an intersection of its line with the line segment $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$, and anything outside of this quadrilateral will have the intersection with $\mathbb{S}^1$.
Formally, these four lines are given by the equations:
$$y = 1 - 2x, y = 2x - 1, y = 2x + 1, y = -2x - 1.$$
I have written them in the order: first quadrant, fourth quadrant, second quadrant, third quadrant.
So, it means that if for a point $\left( x, y \right) \in B_1$ with $0 \leq y \leq 1 - 2x$ or $2x - 1 \leq y \leq 0$ or $0 \leq y \leq 2x + 1$ or $-2x - 1 \leq y \leq 0$, we would expect the line joining $\left( x, y \right)$ with $\left( \frac{1}{2}, 0 \right)$ or $\left( - \frac{1}{2}, 0 \right)$ (depending on the sign of $x$) to intersect the line segment $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$.
First, we see for $x \geq 0$. The line joining $\left( x, y \right)$ with $\left( \frac{1}{2}, 0 \right)$ is given by the equation:
$$y' = \left( \dfrac{y}{x - \frac{1}{2}} \right) \left( x' - \dfrac{1}{2} \right).$$
Here, we assume that $x \neq \frac{1}{2}$. We will consider this case later. Now, whenever $0 \leq y \leq 1 - 2x$ or $2x - 1 \leq y \leq 0$, this line intersects with $\left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right]$. The point of intersection will be $\left( 0, \frac{-y}{2x - 1} \right)$. Thus, for the points $\left( x, y \right)$ in the right region of the orange quadrilateral, we get a point $\left( 0, \frac{-y}{2x - 1} \right)$ in the theta space.
Similar computation would tell us that the points $\left( x, y \right)$ in the left half region of the orange quadrilateral should be mapped to $\left( 0, \frac{y}{2x + 1} \right)$.
Now, let us see what happens when $x = \frac{1}{2}$. Since $\left( frac{1}{2}, 0 \right)$ is not included in $B_1$, any point of the form $\left( \frac{1}{2}, y \right)$ will lie outside the orange quadrilateral. Therefore, we now look at the intersection of the corresponding line with $\mathbb{S}^1$. Since on the line $x = \frac{1}{2}$, the intersection will be at $\left( \frac{1}{2}, \pm \frac{\sqrt{3}}{2} \right)$, depending on the sign of $y$.
Similar things can be said about $x = - \frac{1}{2}$. Seethe figure for a better understanding. Now, let us only look at the case when $x \neq \pm \frac{1}{2}$ and $\left( x, y \right)$ lies outside the orange region. Let us first consider the right half plane. Now, we have to look at the intersection of the line $y' = \left( \dfrac{y}{x - \frac{1}{2}} \right) \left( x' - \dfrac{1}{2} \right)$ with $\mathbb{S}^1$. Upon calculation, we get a quadratic equation in $\left( x' - \dfrac{1}{2} \right)$, whose solution(s) are given by
$$x' - \dfrac{1}{2} = \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}$$
Now, $x' \geq 0$ should hold (otherwise the line would first intersect the segment $\left\lbrace 0 \right\rbrace \times [-1, 1]$, which is not desired for now). Therefore, for the purple points, we get the point
$$\left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)} + \dfrac{1}{2}, \left( \dfrac{y}{x - \frac{1}{2}} \right) \left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)} \right) \right)$$
Similar computations can be done for purple points in the left half part of $B_1$. So, the function $f: B_1 \rightarrow B_1$ which we wanted to construct is as follows:
$$f \left( x, y \right) = \begin{cases}
\left( 0, \frac{-y}{2x - 1} \right), & x \geq 0 \text{ and } \left( 0 \leq y \leq 1 - 2x \text{ or } 2x - 1 \leq y \leq 0 \right) \\
\left( 0, \frac{y}{2x - 1} \right), & x \leq 1 \text{ and } \left( 0 \leq y \leq 2x + 1 \text{ or } -2x - 1 \leq y \leq 0 \right) \\
\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right), & x = \frac{1}{2} \text{ and } y > 0 \\
\left( \frac{1}{2}, - \frac{\sqrt{3}}{2} \right), & x = \frac{1}{2} \text{ and } y < 0 \\
\left( - \frac{1}{2}, \frac{\sqrt{3}}{2} \right), & x = - \frac{1}{2} \text{ and } y > 0 \\
\left( - \frac{1}{2}, - \frac{\sqrt{3}}{2} \right), & x = - \frac{1}{2} \text{ and } y < 0 \\
\left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)} + \dfrac{1}{2}, \left( \dfrac{y}{x - \frac{1}{2}} \right) \left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x - \frac{1}{2}} \right)^2 \right)} \right) \right), & x \geq 0 \text{ and } x \neq \frac{1}{2} \text{ and } y > 1 - 2x \text{ and } y < 2x - 1 \\
\left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x + \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x + \frac{1}{2}} \right)^2 \right)} - \dfrac{1}{2}, \left( \dfrac{y}{x + \frac{1}{2}} \right) \left( \dfrac{-1 \pm \sqrt{1 + 3 \left( 1 + \left( \frac{y}{x + \frac{1}{2}} \right)^2 \right)}}{2 \left( 1 + \left( \frac{y}{x + \frac{1}{2}} \right)^2 \right)} \right) \right), & \text{otherwise}
\end{cases}$$
As you can see, writing the function explicitly is way too tedious and lengthy. However, with this analytical expression, one can satisfy themselves about the continuity of $f$.
As for your question about homotopy between the identity function and this particular function we have constructed, we look at the following. While constructing this function, we essentially saw lines and "moved" each point along a certain line until we reached the theta space. So, a natural way to deform the disc (with two points removed) into the theta space is that at the beginning, stay where you are and then start moving along the lines we have considered. We know that then each of these lines (segments) would then start at a point in $B_1$ and end at a point in $\mathbb{S}^1 \cup \left( \left\lbrace 0 \right\rbrace \times \left[ -1, 1 \right] \right)$. Use this intuition to define the homotopy as $H: B_1 \times \left[ 0, 1 \right] \rightarrow B_1$,
$$H \left( \textbf{x}, t \right) = \left( 1 - t \right) \textbf{x} + t f \left( \textbf{x} \right).$$
Now, we can easily verify that this is indeed a homotopy between the identity map and the map $f$, we have constructed.
Best Answer
For part 1, if you remove a point from the arcs A and B, can we deform the the line segments to anything?
For part 2, just remove the point from arc A and see what space you get after deformation.