Why is the identity map of the circle not straight-line homotopic to a constant map? According to the book 'Introduction to Topology. Pure and Applied' by C Adams & R Franzosa :

$\boldsymbol{\sf THEOREM\ 9.9.}$ A circle function $f : S^1 \rightarrow  S^1$ has degree $0$ if and only if $f$ extends to a continuous function on the disk $D$ (that is, if and only if there exists a continuous function $F : D \rightarrow  S^1$ such that $F (x) = f (x)$ for all $x\in S^1$).

I read the proof and it seems cool! But, the contradiction of the theorem itself is that : If $f$ is the identity function, i.e. if $f (\theta) = \theta$ (degree $1$), it still is extendable on the Disk and a straight-line homotopy $F(\theta, t) = (1-t) f (\theta)$ deforms $f (\theta)$ to a constant map which has degree $0$. Contradiction? 
PS - Please let me know to add the proof below if it's necessary but I think the proof is irrelevant.  

Proof. Throughout the proof, we represent the points in the disk with polar coordinates $(r,\theta)$.
$\quad$ First, assume that $f$ extends to a continuous function $F:D\to S^1$. Define a function $G:S^1\times I\to S^1$ by $G(\theta,t)=F(t,\theta)$. Since $F$ is continuous, so is $G$. Therefore $G$ is homotopy between the circle functions $G|_{S^1\times\{0\}}$ and $G|_{S^1\times\{1\}}$, implying that these functions have the same degree. The function $G|_{S^1\times\{0\}}$ is given by $$G|_{S^1\times\{0\}}(\theta)=G(\theta,0)=F(0,\theta)=F(0,0).$$ It follows that $G|_{S^1\times\{0\}}$ is a constant function and therefore has degree $0$. Thus, $G|_{S^1\times\{1\}}$ also has degree $0$. It is straightforward to see that $G|_{S^1\times\{1\}}$ is equal to $f$, implying that $f$ has degree $0$, as desired.
$\quad$ Now assume that $f$ has degree $0$. Therefore there exists a homotopy $G:S^1\times I\to S^1$ such that $G(\theta,0)=c_0(\theta)$ and $G(\theta,1)=f(\theta)$, where $c_0:S^1\to S^1$ is the constant function sending each point $\theta\in S^1$ to the point $0\in S^1$. Define $F:D\to S^1$ by $F(r,\theta)=G(\theta,r)$. Since $G(\theta,r)$ is constant in $\theta$ when $r=0$, it follows that $F$ is well defined at $r=0$ and therefore is defined as a function on $D$. Furthermore, $G$ being continuous implies that $F$ is  as well. Finally, $F(1,\theta)=G(\theta,1)=f(\theta)$, implying that $F$ is an extension of $f$. $\tag*{${\blacksquare}{}$}$

 A: If $f:\mathbb{S}^1 \to \mathbb{S}^1$ were homotopic to a constant map, you would need the homotopy to stay within $\mathbb{S}^1$, $$F:\mathbb{S}^1\times I \to \mathbb{S}^1$$ But it doesn't: $$|(1-t)f(\theta)|=|1-t||f(\theta)|=|1-t|\neq 1, \text{ for every } t\neq 0.$$
So you have the identity on the disk homotopic to a constant (what you have proven via $F:D\times I \to D$), but the identity on the sphere cannot be homotopic to a constant via your homotopy via restriction of your homotopy (because $F(\mathbb{S}^1\times I)\nsubseteq \mathbb{S}^1$, as we have shown above).  
EDIT: It is true that $F:D\times I \to D$ given by $F(x,t)=tx_0+(1-t)x$ is a straightline homotopy between the identity function $id:D\to D$ and the constant any point $x_0$ of the disk, $c:D\to D$, $c: x\mapsto x_0$. Just check the definition for this homotopy.
Your confusion arises because the degree of a map is defined for maps $f:\mathbb{S}^n\to \mathbb{S}^n$. It is precisely the integer $d$ such that $H_n(f):H_n(\mathbb{S}^n)\to H_n(\mathbb{S}^n)$ is given by $H_n(f)[x]=d[x]$, where $[x]$ is a fixed generator for $H_n(\mathbb{S}^n)\cong \mathbb{Z}$, being your case $n=1$. Sorry for the mess.
Also, note that there is no contradiction in the theorem because $G$ is being restricted to $\mathbb{S}^1\times \{0\}$, which is homeomorphic to $\mathbb{S}^1$ and therefore it makes sense to talk about degree.
