Understanding the DTFT
From the last highlighted paragraph:


It's perhaps easier to relate a unit delay in the z and s domains; thus \$e^{-sT}\$ delays a continuous signal by \$\small T\$ secs, and \$z^{-1}\$ is the unit delay function in the z-domain, which delays a discrete signal by one sampling increment. Hence \$e^{-sT}\leftrightarrow z^{-1}\$, or \$e^{sT}\leftrightarrow z^{1}=z\$ is often a more convenient form.
To go from the s-domain to the frequency domain we use \$s \rightarrow j\omega\$. Using the equivalence in 1., above we may go from the z-domain to frequency domain by: \$\small z\rightarrow e^{st} \rightarrow e^{j\omega T} = cos(\omega T) +j\:sin(\omega T)\$. Now, \$\small \omega T\$ is an angle, which is proportional to \$\small \omega\$, and it's convenient to denote this angle, \$\small \Omega\$, and think of \$\small e^{j\Omega}=cos\:\Omega\:+\:j\:sin\:\Omega\$ as a vector that rotates counter-clockwise from zero radians as frequency increases from \$\small\omega=0\$ rad/sec.
Now consider an s-plane root, \$\small s=-\alpha\$. This would transform to a z-plane root: \$\small z \rightarrow e^{-\alpha T}\$, which is real, positive and less than unity in magnitude.
Moving on, a complex root in the s-plane: \$\small s=\:-\alpha\:+ j\:\omega\$ transforms to \$\small z \rightarrow e^{-\alpha T\:+ j\:\omega T}=\: e^{-\alpha T}\:\small (cos\:\Omega\:+\:j\:sin\Omega)\$, which is a counter-clockwise rotating vector with radius: \$\small e^{-\alpha T}\$. The conjugate root would rotate clockwise with the same radius. Note that the radius is \$\small \lt 1\$, i.e. the vectors are inside the unit circle.