This (elementary and perfectly standard) example might help show the power of sheaves with non-constant coefficients:
First, think about the circle $S^1$. Suppose you want to understand (real) line bundles on the circle. You can certainly cover the circle with two open contractible subsets $U_1$ and $U_2$ (which you can take to be the complements of the north and south poles), and we know that any line bundle on a contractible space is trivial. So if you've got a line bundle $L$ over $S^1$, you can restrict it to either $U_i$ and get a trivial bundle $L_i$. $L$ is built from these $L_i$ and the way they they are patched together over $U_1\cap U_2$.
Now what does it mean to patch the $L_i$ together over $U_{12}=U_1\cap U_2$? It means choosing an isomorphism $L_1|U_{12}\rightarrow L_2|U_{12}$. For any $x\in U_{12}$, the restriction of this isomorphism to the fiber $L_x$ over $x$ is an isomorphism between 1-dimensional vector spaces, and so (after choosing bases) can be identified with an element of ${\bf R}^*$ (the non-zero reals). Therefore your patching consists of a continuous map
$$U_{12}\rightarrow {\mathbb R}^*$$
which is to say, a Cech 1-cocycle for the sheaf of continuous ${\bf R}^{*}$-valued functions.
Now of course you could build a line bundle in some other way, say by starting with two different contractible sets $U_1$ and $U_2$. When do two sets of patching data give isomorphic line bundles? A little thought reveals that the answer is: When and only when the corresponding cocycles give the same class in
$$H^1(S^1,G^{*})$$
with $ G^{*} $ being the sheaf of continuous ${\bf R}^*$-valued functions.
Therefore line bundles are classified by $H^1(S^1,G^{*})$. Now consider the exact sequence of sheaves
$$0 \rightarrow G \rightarrow G^*\rightarrow {\bf Z}/2{\bf Z}\rightarrow 0$$
where $G$ is the sheaf of continuous ${\bf R}$ valued functions, and the map on the left is exponentiation. Follow the long exact sequence of cohomology, use the fact that $G$ is acyclic, and conclude that $H^1(S^1,G^*)=H^1(S^1,{\bf Z}/2{\bf Z})={\bf Z}/2{\bf Z}$. In other words, there are exactly two real line bundles over $S^1$ --- and indeed there are: the cylinder and the Mobius strip.
Exercise: Do a similar calculation for ${\bf CP}^1$ (the Riemann sphere). Conclude that the set of (complex) line bundles is in one-one correspondence with $H^2({\bf CP}^1,{\bf Z})={\bf Z}$.
This answers address the part of the question asking what the divisors above correspond to in number theory and ring theory. I also include some vague remarks on the naming.
The relevance of Dedekind domains in the geometric context was already mentioned in a comment. And, the notion of Dedekind domains in some sense 'comes from' number theory.
The rings of algebraic integers of algebraic number fields are important examples of Dedekind domains and the historical root for the development.
Rings of algebraic integers are not necessarily unique factorization domains, and one tried to remedy this lack of uniqueness of factorization into irreducible elements by passing to a different (larger) structure. A structure consisting of 'ideal elements' that would not lack the nice property of having unique factorizations into irreducibles.
Now, todays ideals are in some sense these 'ideal elements' for rings of algebraic integers and more generally Dedekind domains.
(One way to define the notion Dedekind domain is: every non-zero ideal is a product of prime ideals.)
Yet, besides the approach via ideals there was also a different/competing approach, namely that of establishing a divisor theory (orig. Divisorentheorie) promoted by Kronecker.
(When I searched for a reference to link to, almost first thing I found was a related MO question, how practical!)
In the number theoretic context today this approach and name is not very wide spread, but might explain the name. And, it is still preserved in more general contexts (cf. below).
But, it is essentially equivalent to considering ideals. And, as mentioned in one of the comments (in suitable circumstances) the fractional ideals correspond to the divisors (in the geometric context).
As said above for a Dedekind domain all non-zero ideals are in a unique way products of prime ideals, so $I= \prod_P P^{v_p}$ where $P$ runs through the non-zero prime ideals and $v_p$ are nonnegative integers all but finitely many $0$. If one allows $v_p$ to be negative too, one gets the fractional ideals.
The link is perhaps even better seen when generalizing a bit, and here the standard terminology even includes words resembling 'divisor'.
Another way to define Dedekind domain would be to say it is a noetherian, integrally closed domain of dimension $1$.
Now, here is a generalization of Dedekind domains: Krull domain.
A way to define what a Krull domain is, is to say that it is a completely integrally closed domain and a v-noetherian domain; where v-noetherian means ACC only on divisorial ideals that is those for which $(I^{-1})^{-1} = I$, in particular v-noetherian is weaker than noetherian; and where completely integerally closed is a condition that is in principle stronger than integrally closed but for noetherian domains it is the same.
So, since Dedekind domains are noetherian and integrally closed they are completely integrally closed, and since they are noetherian they are v-noetherian. Thus, they are Krull domains.
More generally, every integrally closed and noetherian domain is Krull (without a condition on dimension!).
In particular, Krull domains have the property that the localization of the domain at every height one prime ideal is a DVR, and with an extra condition one could take this property as definition, too.
Now, the problem arises that also in Krull domains one would like to have 'ideal elements' or a 'divisor theory'. Yet, the usual ring ideals are in this case not the 'ideal elements'. One needs to consider a different type of ideals namely the divisorial ideals.
It is then true that every divisorial ideal can be uniquely factorized as a product of prime divisorial ideals (in perhaps more classical terms these are the prime ideals of height one). A key-point is that principal ideals are divisorial ideals, and what one cares about mainly is to have a substitute of unique factorization for the ring elements, so principal ideals.
The product used for this factorization is not the usual product of ideals, but rather the product as divisorial ideals that is one forms $((IJ)^{-1})^{-1}$ to get the product of $I$ and $J$.
One can still generalize this more to commutative semigroups with identity and cancellation law. Namely, a semigroup $S$ with a divisor theory is one for which there exists a free commutative semigroup $F$ and a homomorphism $f:S \to F$ such that $a|b$ if and only if $f(a)|f(b)$ for all $a,b \in S$ [one implication is trivial as $f$ is a homomorphism, but the other one is important]. In other words, the arithmetic (or divisibility relations in $S$) are governed directly by the ones in a free semigroup (so a factorial/unique factorization one).
Side note: the $f$ above is not strictly speaking a divisor theory as there is a minimality condition omitted, but still one can rigorously take this as definition; let us assume for simplicity we have in addition this suitable minimality condition (whose definition we omit).
Then we have an $S$ which we care about and an $F$ which is easy to understand (consisting of the 'divisors'). So, the divisors of $S$ control or govern the divisibility relations in $S$. To be a bit more precise, in $F$ itself we only have what corresponds to the positive divisors (that is no negative 'coefficient'). However, to make the link more direct, one can (and does, e.g., for defining the class group) instead consider the quotient group, which consists precisely of all (formal) products $\prod_p p^{s_p}$ where $p$ runs through the prime elements of $F$ the $s_p$ are integers (all but finitely many $0$). [The sole difference is that here multiplicative instead of additive notation is used.]
What is the link of these semigroups to domains: a domain is Krull if and only if its multiplicative semigroup is a semigroup with divisor theory.
In fact, one can take the semigroup of fractional ideals for the $F$.
And, since a Dedekind domain is Krull its multiplicative monoid is also a semigroup with divisor theory and in this case one can take $F$ to be the (usual) ideals.
In view of the above, if one completely reduces down to the most classical number theory situation of the integers or natural numbers then everything 'collapses' and the positive divisors (so no negative 'coefficient') 'are' just the elements of the original structure, except for zero and 'forgetting the sign' for the integers. [Omitting the positivity condition, the totality of divisors would correspond to the respective quotient group, i.e., the positive rationals with multiplication (one allows positive and negative exponents in the 'prime factorization')]
In the classical situation of algebraic number theory (ring of integers of an algebraic number field) and more generally Dedekind domains the positive divisors correspond to the ideals and the divisors to the fractional ideals.
And, one could say, these divisors control or govern the divisibility relations (in an efficient way). This is also true for the integers; what one omits is the sign which is in fact irrelevant for divisibility.
So, the divisors (in the geometric context) are closely linked (in certain cases the same) as objects that in number and ring theoretic contexts describe or govern the arithmetic, i.e. divisibility relations.
Best Answer
Disclaimer. I am no expert at all in algebraic geometry. Therefore much of the following will be oversimplified or maybe even simply wrong. You are still invited to improve it.
EDIT. There is a paper by Totaro where he shows that the cycle class map from the Chow group to singular cohomology factors as $$CH^*(X)\longrightarrow MU^*(X)\otimes_{MU^*}\mathbb Z\longrightarrow H^*(X,\mathbb Z)\;.$$ Here $MU^*$ denotes complex cobordism, as explained below. The fact that Totaro does not construct a map through $MU^*(X)$ seems to indicate that in general, equivalence of cycles can be coarser than complex cobordism. I have not checked the implications for divisors, that is, for $CH^1(X)$.
Let $L\to V$ be a line bundle over a complex variety, let $s$ be a meromorphic section of $L$, and let $D$ be the associated divisor. For simplicity, we assume that $s$ is an algebraic section and that $s$ meets $0$ transversely, so all zeros have multiplicity one. Then $D$ `is' the subvariety $s^{-1}(0)$. If $V$ was smooth, then $D$ is a subvariety of complex codimension $1$, which we regard as a cobordism $2$-cocycle. Its normal bundle is a complex line bundle.
A linear equivalence between two divisors $D_0$, $D_1$ is given by a meromorphic function $f$ such that $D_a=f^{-1}(a)$ for $a=0$, $1$. Consider the graph $\Gamma$ of $f$ in $V\times\mathbb P^1$. If we are still over $\mathbb C$ and everything is sufficiently regular, for a generic real smooth curve $c\colon[0,1]\to\mathbb P^1$ from $0$ to $1$ in $\mathbb P^1$ (now with analytic topology), the (transverse) intersection $W$ of $\Gamma$ and $V\times\operatorname{im}(c)\cong V\times[0,1]$ is then a cobordism between $D_0\times\{0\}$ and $D_1\times\{1\}$. The normal bundle of $W$ in $V\times[0,1]$ naturally carries the structure of a topological complex line bundle, which restricts to the normal bundles of $D_0$, $D_1$ in $V$.
To see that we get a map from the divisor class group to complex cobordism, we recall cocycles and relations. A $k$-cocycle in a manifold $M$ is a submanifold $D\subset M\times\mathbb R^\ell$ of real codimension $(k+\ell)$ together with a complex structure on its normal bundle. If one projects down to $M$, the image may become singular. One can stabilise in $\ell$ by taking $D\times\{c\}\subset M\times\mathbb R^{\ell+m}$.
A cobordism between two $K$-cocycles, represented by submanifolds $D_0$, $D_1\subset M\times\mathbb R^\ell$ is a submanifold $W\subset M\times\mathbb R^\ell\times[0,1]$ such that $\partial W=D_0\times\{0\}\sqcup D_1\times\{1\}$, together with a complex structure on the normal bundle that restricts to the given complex structures of the normal bundles of $D_0$, $D_1\subset M\times\mathbb R^\ell$.
This shows that sufficiently smooth divisors give rise to cocycles, and linear equivalence implies cobordism. It seems that this construction extends to Chow groups, and I am sure it is described somewhere with greater care. It is not entirely clear to me though how to deal with singular subvarieties.
To push the analogy further, to each algebraic line bundle one associated a divisor class $[D]$. This can be regarded as a universal first Chern class, with values in the Chow ring. On the topological side, there is a first Chern class with values in the complex cobordism ring, and it corresponds to $[D]$ under the map above. By Quillen, this class is universal for complex oriented multiplicative cohomology theories.
Conversely, in algebraic geometry, a divisor defines a line bundle. If we can choose $\ell=0$ above, then the Pontryagin-Thom construction identifies a complex $2$-cocycle in $M$ with a homotopy class of maps $M\to\mathbb P^\infty$, and $\mathbb P^\infty$ is also the classifying space for (topological) complex line bundles. There is a direct construction: the normal bundle of $D$ comes with a complex structure. Pull it back to a tubular neighbourhood $\pi\colon U\to D$ of $D$, then $\pi^*\nu\to U$ has a tautological section, which can be used to glue it to trivial bundle over $M\setminus D$, giving a complex line bundle $L\to M$. This construction is unstable however, that is, it does not work for higher values of $\ell$.