Both results are actually equivalent. You can prove one from the other.
Regarding the first result:
Let $\mathcal{C}$ be a class of subsets of $\Omega$ closed under finite intersections and containing $\Omega$. Let $\mathcal{B}$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $\mathcal{B} = \sigma ( \mathcal{C})$.
Some books call it "Monotone Class Theorem", although this is not the most usual naming.
A class having $\Omega$, closed under increasing limits and by difference is called a "Dynkin $\lambda$ system". A non-empty class closed under finite intersections is called a "Dynkin $\pi$ system".
The result above can be divided in two results
1.a. A $\lambda$ system which is also a $\pi$ system is a $\sigma$-algebra.
1.b. Given a $\pi$ system, the smallest $\lambda$ system containing it is also a $\pi$ system.
Some books call result 1.a (or result 1.b) "Dynkin $\pi$-$\lambda$ Theorem.
Some quick references is
https://en.wikipedia.org/wiki/Dynkin_system
The second result
Let $G$ be an algebra of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = M(G)$.
Where a monotone class in a set $R$ is a collection $M$ of subsets of $R$ which contains $R$ and is closed under countable monotone unions and intersections.
is usually called "Monotone Class Lemma" (or theorem) you can find it in books like Folland's Real Analysis or Halmos' Measure Theory. In fact, Halmos presents a version of this result for $\sigma$-rings.
Let $G$ be ring of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-ring generated by $G$.
Let us prove that the results are equivalent
Result 1: Let $\mathcal{C}$ be a class of subsets of $\Omega$ closed under finite intersections and containing $\Omega$. Let $L(\mathcal{C})$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $L(\mathcal{C}) = \sigma ( \mathcal{C})$.
Result 2: Let $G$ be an algebra of sets and define $M(G)$ to be the smallest monotone class containing $G$. Then $M(G)$ is precisely the $\sigma$-algebra generated by $G$, i.e. $\sigma(G) = M(G)$.
Where a monotone class in a set $R$ is a collection $M$ of subsets of $R$ which contains $R$ and is closed under countable monotone unions and intersections.
Proof:
(2 $\Rightarrow$ 1). Note that any class containing $\mathcal{C}$ which is closed under increasing limits and by difference is close by complement because $\Omega \in \mathcal{C}$, and so it is also closed by decreasing limits. So it is closed under countable monotone unions and intersections. It means: any class containing $\mathcal{C}$ which is closed under increasing limits and by difference is a monotone class.
Note also that any class containing $\mathcal{C}$ which is closed under increasing limits and by difference contains $A(\mathcal{C})$ the algebra generated by $\mathcal{C}$.
Then using Result 2 we have
$$ \sigma(\mathcal{C}) = \sigma(A(\mathcal{C})) = M(A(\mathcal{C})) \subseteq L(A(\mathcal{C}))=L(\mathcal{C}) $$
Since $\sigma(\mathcal{C})$ is a class containing $\mathcal{C}$ which is closed under increasing limits and by difference, we have $L(\mathcal{C}) \subseteq \sigma(\mathcal{C})$, so $L(\mathcal{C}) = \sigma(\mathcal{C})$.
(1 $\Rightarrow$ 2). First let us prove that $M(G)$ is a class containing $G$ which is closed under increasing limits and by difference. Since $M(G)$ is monotone, we have that $M(G)$ is closed under increasing limits.
Now, for each $E\in M(G)$, define
$$M_E=\{ F \in M(G) : E\setminus F , F \setminus E \in M(G) \}$$
Since $M(G)$ is a monotone class, $M_E$ is a monotone class. Moreover, if $E\in G$ then for all $F \in G$, $F\in M_E$, because $G$ is an algebra. So, if $E\in G$, $G \subset M_E$. So, if $E\in G$, $M(G) \subset M_E$. It means that for all $E\in G$, and all $F \in M(G)$, $F \in M_E$. So, for all $E\in G$, and all $F \in M(G)$, $E \in M_F$. So, for all $F \in M(G)$, $G \subset M_F$, but since
$M_F$ is a monotone class, we have, for all $F \in M(G)$, $M(G)\subset M_F$. But that means that $M(G)$ is closed by differences.
So we proved that $M(G)$ is a class containing $G$ which is closed under increasing limits and by difference.
So by Result 1, $$\sigma(G)=L(G) \subseteq M(G)$$
Since $\sigma(G)$ is a monotone class, we have
$$ M(G) \subseteq \sigma(G)$$
So we have $$\sigma(G)= M(G)$$
Best Answer
If I get things wrong, let me know, please. Exercise 2.5(b) is useful in this exercise.
Now we prove Exercise 3.12. As was noticed by OP, it suffices to prove that $f(\mathscr{P})\subset \mathscr{L}$. Since $\mathscr{L}$ is closed under the disjoint unions, it suffices to prove that $\bigcap_{j=1}^{n}P_{j}\in\mathscr{L}$, where for each $j$ either $P_j\in\mathscr{P}$ or $P_j^c\in\mathscr{P}$. Without loss of generality, we assume that $P_1^c,\cdots P_k^c\in \mathscr{P}$ and $P_{k+1},\cdots, P_n\in\mathscr{P}$. Let $P=P_{k+1}\cap\cdots\cap P_n\in\mathscr{P}$ and $A_j=P_j^c\in\mathscr{P}$ for $1\leq j\le k$, then \begin{align*} \bigcap_{j=1}^{n}P_{j}&=P\cap A_1^c\cap A_2^c\cap\cdots\cap A_k^c=P\bigcap\left(A_1\cup A_2\cup \cdots \cup A_k\right)^c\\ &=P\setminus \left(P\cap\left(A_1\cup A_2\cup \cdots \cup A_k\right)\right)\\ &=P\setminus\left((P\cap A_1)\cup(P\cap A_2)\cup\cdots\cup(P\cap A_k)\right). \end{align*} Since $\mathscr{L}$ is closed under the formation of proper differences, it suffices to show that $$(P\cap A_1)\cup(P\cap A_2)\cup\cdots\cup(P\cap A_k)\in\mathscr{L}.\tag{1}$$ The key point is to write the above expression as a dijoint union of sets in $\mathscr{L}$. Indeed, we have \begin{align*} &(P\cap A_1)\cup(P\cap A_2)\cup\cdots\cup(P\cap A_k)\\&=(P\cap A_1)\cup((P\cap A_2)\setminus(P\cap A_1\cap A_2))\cup\cdots\cup((P\cap A_k)\setminus(P\cap A_1\cap A_2\cap\cdots\cap A_k)). \end{align*} Now since $\mathscr{P}$ is closed under finite intersections, $\mathscr{L}$ is closed under the formation of proper differences and the disjoint unions, we get $(1)$. Therefore, $\bigcap_{j=1}^{n}P_{j}\in\mathscr{L}$ and the proof is completed.