For a filter $F$ in a boolean algebra $B$, they define
$$ x \sim_F y \text{ iff for some } f\in F, x \wedge f = y \wedge f$$
They go on to prove that this is a equivalence relation and then define the quotient algebra $B/F$ where $\wedge, \vee, *$ applied to the equivalence classes are just the same operators applied to any two representative elements.
My question is, why is this the definition for $B/F$? Clearly any equivalence relation will allow a forming a quotient algebra, and there many equivalences relations on $B$; and though I can't think of any non-trivial one, there are probably some other equivalence relations on $B$ that "depend" somehow on $F$.
Right after the definition, they present some kind of analogue to the first homomorphism theorem for groups, so if $f: B_1 \to B_2$, then $f[B_1] \cong B_1/F$, where $F= \{ x \in B_1 \mid f(x)=1 \}$. I think this is a property that should be satisfied for any definition of quotient algebra, but again, maybe there are some other definitions of quotient algebra that satisfy it too.
Sure this definition has other nice properties like $B/F \cong 2$ whenever $F$ is an ultrafilter, etc. But all this seems like a coincidence for me right now. So my question is to justify the naturality of this construction.
Although you mentioned the theorem, you seem to have missed the significance of the fact that if $f:B_1\to B_2$ is a homomorphism, then $\{x\in B_1:f(x)=1_{B_2}\}$ is a filter in $B_1$, and $f[B_1]\cong B_1/F$. The theorem basically says that if we want quotient algebras of $B_1$ to be homomorphic images of $B_1$, this is the only way to define them: the homomorphism determines the filter, and vice versa.
The equivalence relation is in any case a very natural one once you realize that one can think of a filter $F$ on a Boolean algebra $B$ as singling out a class of ‘large’ elements of $B$. This is perhaps clearest in the case of power set algebras. A simple example is the cofinite filter on $\wp(\Bbb N)$ consisting of all subsets of $\Bbb N$ with finite complements: being cofinite is clearly a notion of largeness. With respect to this notion the finite sets are small, and the infinite sets that are not cofinite are neither large nor small.
Defining $x\sim_Fy$ iff $x\land f=y\land f$ for some $f\in F$ is in a sense saying that $x$ and $y$ agree on a large element of $B$. Here again the power set algebras provide the intuition. For instance, if $\mathscr{F}$ is the cofinite filter on $\wp(\Bbb N)$, and $X,Y\subseteq\Bbb N$, then $X\sim_{\mathscr{F}}Y$ iff there is an $F\in\mathscr{F}$ such that $X\cap F=Y\cap F$, i.e., iff there is a finite $A\subseteq\Bbb N$ such that $X\setminus A=Y\setminus A$. This is sometimes expressed by saying that $X$ and $Y$ agree almost everywhere modulo finite sets: apart from a finite subset of $\Bbb N$, they have the same elements. Indeed, $X\sim_{\mathscr{F}}Y$ iff $X\mathrel{\triangle}Y$ is finite.