Simply connected topological space, a detail in the definition I do not understand here in Definition and equivalent formulations paragraph where do we have that point in question by the mere fact that there exists a continuous map $F:D^2\to X.$ I.e. which point is obtained by this definition: by existence of this continuous $F$ ?
 A: Let $f:\Bbb S^1\to X$ be null-homotopic, say $F:\Bbb S^1\times [0,1]\to X$ with $F(\bullet,0)=f$ and $F(\bullet,1)=c_{x_0}$, where $c_{x_0}$ is the constant map based at $x_0\in X$. Then, $g:\Bbb D^2\to X$ defined by $$g(z)=\begin{cases} x_0 &\text{ if }0\leq ||z||\leq \frac{1}{2},\\ F\left (\frac{z}{||z||},2-2||z||\right) &\text{ if } \frac{1}{2}\leq ||z||\leq 1.\end{cases}$$ is continuous by pasting lemma and $g(z)=F(z,0)=f(z)$, i.e. $g$ extends $f$.

Conversely, suppose $f:\Bbb S^1\to X$ is a map and $g:\Bbb D^2\to X$ extends $f$ i.e. $g\big|_{\Bbb S^1}=f$.  Define, $F:\Bbb S^1\times [0,1]\to X$ as $$F(z,t)=g\left((1-t)z+tz_0\right), \text{ where }z_0\in \Bbb S^1\text{ be given fixed point.}$$ Notice that $F(z,1)=g(z_0)=f(z_0)$ for all $z\in\Bbb S^1$. Hence, $F:f\simeq c_{f(z_0)}$.

In the first part, the construction of $g$ can be done in the
following way also. Let $\pi:\Bbb S^1\times [0,1]\to \Bbb D^2$  be the
map $\pi(z,t)=(1-t)z$. Note that $\pi$ is continuous, closed, and
subjective, so $\pi$ is a quotient map. Now, for any $F:\Bbb S^1\times
 [0,1]\to X$ with $F(\bullet,0)=f$ and $F(\bullet,1)=c_{x_0}$ we have
$F$ is constant on $\Bbb S^1\times \{1\}$, so $F$ induces  via the quotient map $\pi$, a continuous map $g:\Bbb D^2\to X$ that is an extension of $f$.

That is to say, we pinch the top-circle $\Bbb S^1\times 1$ of the cylinder $\Bbb S^1\times [0,1]$, to make it a cone over the bottom circle $\Bbb S^1\times 0$, and then we use the fact that cone over $\Bbb S^1$ is homeomorphic to $\Bbb D^2$.
