Here's my take on it, those conditions are wrong when it comes to defining if the feedback is positive/negative, so
1-) β being negative causes 180 degree phase shift so there is positive feedback
For this block diagram, if \$A\$ is considered the plant, the feedback is negative if \$\beta>0\$ (doesn't mean it will stabilize the plant/loop or anything). If the plant is something after \$A\$ (an unitary gain block?), then the feedback is negative for this block diagram if \$A\beta>0\$.
2-) Condition for positive feedback: |1 + β*A| < 1 . β being negative alone is not enough for positive feedback saturation.
For the loop to oscillate, you need the meet the Barkhausen stability criterion (and only simple poles), where \$|A\beta|\ = 1\$ and also (in this case due to the negative sum junction) you must have either \$A < 0\$ or \$\beta < 0\$.
Also, the book states three things in this page that are particularly confusing.
In the Eq. 9.3 the feedback voltage \$V_f\$ is presented to the input circuit in subtractive fashion. The denominator \$|1+A\beta| > 1\$ and the feedback is negative. Equation 9.5 then shows that \$|A'|<|A|\$ and the gain of the system with feedback is less than the internal amplifier gain. Thus gain is sacrificed with negative feedback.
I read the paragraph as, if \$|1+A\beta| > 1\$ and feedback is negative (\$\beta > 0\$, since the picture shows it just multiplies the output and goes to a negative sum junction), then \$|A'|<|A|\$. Or logically,
$$|1+A\beta| > 1,~ (\beta>0)^* \Rightarrow |\frac{A}{1+A\beta}| =|A'|<|A|.$$
*Notice this is useless to his proof, removing this premise doesn't break the conclusion. It is probably mentioned as it is common to have negative feedback systems where \$\beta>0, ~A>0\$
If \$A\$ is negative, as is usual in C-E amplifiers, we reverse \$V_f\$ from the \$\beta\$ network, resulting in a positive \$A\beta\$ term in Eq. 9.5 and so retain the negative feedback.
I suppose it mean, if \$A<0\$ we need \$\beta < 0~\$ ("... we reverse Vf from the β network") to keep \$|A'|<|A|\$. Or logically,
$$A<0,~ \beta<0 \rightarrow |1+A\beta| > 1 \Rightarrow |\frac{A}{1+A\beta}|=|A'|<|A|.$$
Finally,
If the phase of \$V_f\$ reverses, as may happen with nonresistive \$\beta\$ networks, the feedback voltage \$V_f\$ becomes additive to \$V_s\$ in Eq. 9.3 and the denominator of Eq. 9.5 shows that \$|1+A\beta|<1\$ and the feedback is positive. The closed-loop gain is \$|A'|>|A|\$ and the gain of the feedback system is greater than the internal gain.
Having \$\beta<0\$ and \$|1+A\beta|<1\$, we have that \$|A'|>|A|\$. Also, logically,
$$ (\beta<0)^*,~ |1+A\beta|<1 \Rightarrow |\frac{A}{1+A\beta}|=|A'|>|A|.$$
*Also useless to the conclusion, but probably used to hint that \$A>0\$.