Question 1: I'm having a hard time interpreting the question "How does this make sense?"—it doesn't give a very good sense of what you're struggling with. Anyway, you say that you are comfortable saying $f_i \in H^0 (U_i,\mathcal{K}_X^\ast)$. But $\mathcal{K}_X^\ast$ is exactly the sheaf of quotients of two non-zero divisors in $\mathcal{O}_X$.
What I think you're having trouble with is how the sheaf $\mathcal{K}_X^\ast / \mathcal{O}_X^\ast$ is constructed. It's a sheafification, which means that every section has a local trivialization. But what is it a sheafification of, exactly? Of the presheaf that maps every affine open $\operatorname{Spec} A$ to the quotient group $K(A)^*/ A^*$, where $K(A)$ is the fraction field—so $K(A)^\ast$ is just the set of quotients of two non-zero-divisors.
(Note: I think that what I wrote there is technically correct, though it's not as simple as I'm suggesting to actually define this presheaf. One needs to work with $\mathcal{K}_X$ as the sheafification of $U\mapsto \mathcal{O}_X(U)[S(U)^{-1}]$, where $S(U)$ is the collection of elements of $\mathcal{O}_X(U)$ which are non-zero-divisors at every stalk in $U$. This is important for ensuring that the representatives $f_i$ actually restrict to quotients of regular elements.)
$\mathcal{K}_X^\ast / \mathcal{O}_X^\ast$ doesn't necessarily look like that on a random affine open. But for any section $s$, $X$ will have a covering by affine opens $\{U_i\}$ such that $s|_{U_i}$ is of that form.
Question 2: It's not necessarily true that $f_i|_{U_i \cap V_j} \in \mathcal{O}_X (U_i \cap V_j)^*$, or even that $f_i|_{U_i \cap U_j} \in \mathcal{O}_X (U_i \cap U_j)^*$. Remember, the latter is in the fraction field, but, for $f_i$ and $f_j$ to be compatible on $U_i \cap U_j$, they must lie in the same coset of the quotient, which is the same as saying: $f_i|_{U_i \cap U_j} \in f_j|_{U_i \cap U_j} \mathcal{O}(U_i \cap U_j)^\ast$ (plus the same thing with $i$ and $j$ reversed).
If you're having trouble seeing how this interacts with the sheafification, you can think about it entirely on stalks—because the stalk of $\mathcal{K}_X^\ast / \mathcal{O}_X^\ast$ really is a quotient group. And you can show that $f_i|_{U_i \cap U_j} \in f_j|_{U_i \cap U_j} \mathcal{O}(U_i \cap U_j)^\ast$ holds if and only if it holds at each stalk of $U_i \cap U_j$.
The comparison between $f_i$ and $g_j$ is identical; he's just indicating that the condition is the same as for $f_i$ and $f_j$. The reasoning is identical as well.
Question 3: Yes. The quotient map $\mathcal{K}_X^\ast \to \mathcal{K}_X^\ast /\mathcal{O}_X^\ast$ you understand, and $\mathcal{O}_X \cap \mathcal{K}_X^\ast \to \mathcal{K}_X^\ast$ is the inclusion of a subsheaf (the one consisting only of those quotients of non-zero-divisors which already lie in the sheaf $\mathcal{O}_X$, which can be understood in a number of different intuitive ways). And a morphism of sheaves induces a morphism of global sections.
This is all another way of saying: a Cartier divisor is effective if we can choose its representatives $f_i$ to all be regular functions.
For example, on $\mathbb{P}^1_k$, with projective coordinates $x$ and $y$, you could construct an effective Cartier divisor as follows: On the open $x\neq 0$, take the regular function $y/x$. On $y\neq 0$, take the regular function $x/y$. On the intersection $U$, $\mathcal{K}_X(U) = k(x/y)$ and $\mathcal{O}_X(U) = k[x/y,y/x]$. But $x/y$ and $y/x$ are both regular functions, so they both lie in the coset $k[x/y,y/x]^\ast$, hence are compatible.
(If you think of the group of Cartier divisors of $\mathbb{P}^1$ up to equivalence as $\mathbb{Z}$, then the one I just described is $2$.)
