Yes, so Georges is correct. The definition is flawed. The correct definition is that that the order of $f$ is the largest integer $k$ so that you can express $f = g^k h$ in the local ring. There is no reason to expect $h$ to be a unit. (Great question, by the way. I'm embarrassed it took me so long.) See the various examples that @JulianRosen and I offered in the comments.
First, each component of the germ $(A,x)$ is contained in a unique irreducible component of the hypersurface $A$: Let $Z$ be an irreducible component of the germ $(A,x)$. Then $Z$ is an irreducible subset of $A$, and since the irreducible components $A_1, A_2, \dotsc$ of $A$ are maximal irreducible subsets, $Z$ is contained in one component, say $Z \subset A_1$. If $Z$ is contained in another component, say $Z \subset A_2$, then $Z \subset A_1 \cap A_2$. But both germs $(A_1,x)$ and $(A_2,x)$ have distinct irreducible components, which are irreducible components of $(A,x)$. So $Z$ is contained in two different irreducible components of $(A,x)$, which contradicts the maximality of $Z$ with respect to irreducible germs.
This means you get a map
$$\{\, \text{components of } (A,x) \} \to \{\, \text{components of } A\,\}.$$
Clearly, every component of $A$ that contains $x$ is in the image of that map. Since the domain is a finite set, there are only finitely many components of $A$ which contain $x$.
Best Answer
1.An analytic subvariety $A$ is called reducible if $A=A_1\cup A_2$ where $A_1,A_2$ are also analytic subvarieties in $X$,distinct from $A$,if A cannot be represented in this form,it is called irreducible.
2.The sum is defined formally.It's important because we probably can define and research more,such as the degree of the divisor...Otherwise,we just discuss one single irreducible hypersurface sounds a little boring...
By the way,maybe we could imagine the sumgeometrically as the union of the irreducible divisors occurring with non-zero coefficients and "multiplicities" given by the coefficients $a_i$.(Although it maybe hard to imagine...).
When we consider effective(all coefficents of $[Y_i]$ are non-negative) Cartier divisors on schemes,we can indeed add them by taking the subscheme defined locally by the product of defining equations for these two divisors. This indeed give precise geometric meaning to the formal sum!