I answer in hopefully didactic order.
What does the author mean by the intermediate texts exactly?
The intermediate texts after $n$ rounds are the 64-bit quantities $L_n\mathbin\|R_n$, numbering these per the specification of DES. $L_n$ and $R_n$ are the halves.
Why after the eighth round of encryption the two halves must be equal?
I do not know if the two halves must be equal. But we can prove that it is enough that the two halves are equal, which is what the quoted text does.
Ignoring $\mathsf{IP}$ and $\mathsf{IP}^{-1}$, the normal equations of DES encryption are, for $1\le n<16$:
$$\begin{align}
L_n&=R_{n-1}&&&R_n&=L_{n-1}\oplus f(R_{n-1},K_n)\\
L_{16}&=L_{15}\oplus f(R_{15},K_{16})&&&R_{16}&=R_{15}
\end{align}$$
As a consequence of the values of the four weak keys $z$, all bits in register $C$ are identical. Therefore $C$ remains invariant as it rotates during encryption. The same applies for $D$. Hence $K_1=K_2=\ldots=K_{15}=K_{16}$. Thus for $1\le i\le16$ we can replace $f(R_i,K_i)$ by a function $f_z(R_i)$ identical in all the rounds, for fixed $w$. The above DES equations turn into
$$\begin{align}
L_n&=R_{n-1}&&&R_n&=L_{n-1}\oplus f_z(R_{n-1})\\
L_{16}&=L_{15}\oplus f_z(R_{15})&&&R_{16}&=R_{15}
\end{align}$$
If $L_8=R_8$ (that is equals halves after round 8) then, by plugging this into the equations for $n=8$ and $n=9$, we get
$$\begin{align}
L_8&=R_7&&&L_8&=L_7\oplus f_z(R_7)\\
L_9&=L_8&&&R_9&=L_8\oplus f_z(L_8)
\end{align}$$
We plug the top two equations into the bottom ones, and get
$$\begin{align}
L_9&=R_7&&&R_9&=\bigl(L_7\oplus f_z(L_8)\bigr)\oplus f_z(L_8)\\
&&&&&=L_7\oplus\bigl(f_z(L_8)\oplus f_z(L_8)\bigr)\\
&&&&&=L_7\oplus0^{32}\\
&&&&&=L_7
\end{align}$$
We used associativity of $\oplus$, that $f_z$ is a function (and identical in all rounds), that for all 32-bit $X$ it holds $X\oplus X=0^{32}$ (the bitstring of 32 zero bits), which is the neutral for $\oplus$.
Wrapping up, we get $L_7=R_9$ and $R_7=L_9$. Thus "the intermediate texts after seven rounds and after nine rounds of encryption will be equal" within an inversion of left and right that the quotation mistakenly sidesteps.
We plug these results into the equations for $n=7$ and $n=10$, and get
$$\begin{align}
R_9&=R_6&&&L_9&=L_6\oplus f_z(R_6)\\
L_{10}&=R_9&&&R_{10}&=L_9\oplus f_z(R_9)\\
\end{align}$$
After simplification as previously, we get $L_6=R_{10}$ and $R_6=L_{10}$. And so on, up to $L_1=R_{15}$ and $R_1=L_{15}$. For the next step we must take care that the the equations for the last round reverse $L_{16}$ and $R_{16}$ compared to others. We obtain $L_0=L_{16}$ and $R_0=R_{16}$.
The 64-bit plaintext $P$ is such that $\operatorname{IP}(P)=L_0\mathbin\|R_0$ and the 64-bit ciphertext block $C$ is such that $\operatorname{IP}^{-1}(L_{16}\mathbin\|R_{16})=C$. It follows that $P=C$. Call that quantity $m$ and we have $\operatorname{DES}_w(m)=m$, Q.E.D.
Why are there $2^{32}$ fixed points?
There are $2^{32}$ possible values for $L_8$, thus for intermediate text $L_8\mathbin\|R_8$ with $L_8=R_8$. By the above reasoning, each gives a value of $m$ with $\operatorname{DES}_w(m)=m$ for a given weak key $w$. And the values thus obtained must be distinct: DES encryption is deterministic for a given key $w$, thus the same input value can't give different intermediate texts after 8 rounds.
Hence there are at least $2^{32}$ fixed points $m$ for a given weak key $w$. I do not know how many more there might be, for each of the four $w$. However, we can tell that the counts are the same for the all-zero and all-one $w$ (ignoring parity bits); same for the two other weak keys. That follows from DES's complementation property.
How to find fixed points for DES weak keys
One option is to modify some DES encryption code to start at round 8 with $L_8=R_8$. The ciphertext will be a fixed point for the weak key used. We can find $2^{32}$ of these for a given weak key $w$, by using a counter for $L_8$. When that's done for a weak key $w$, we can derive fixed points for the complement of $w$ by using DES's complementation property.
