Is every ring/module the filtered colimit of its coherent/quasicoherent subrings/submodules? What about finitely presented subobjects? What's the intuition behind each case?
Notation. Let $I,J$ be two ideals of commutative ring. If $I=aR,J=bR$, denote the conductor of $J$ into $I$ by $(I:J)$ by $(a:b)$.
Definition. A commutative ring is quasicoherent if for any finite subset $ \left\{ a_1,\dots ,a_n,c \right\} \subset R$ the ideals $a_1R\cap \cdots \cap a_nR$ and $(0:c)$ are finitely generated.
Very generally, any model of a finitary algebraic theory is a filtered colimit of finitely presented objects. For any such object $X$ has a presentation $\langle G \mid R\rangle$, where $G$ is some set of generators and $R$ is some set of relations (and each relation only involves finitely many generators). Now consider the directed poset $P$ consisting of all pairs $(G_0,R_0)$, where $G_0\subseteq G$ and $R_0\subseteq R$ are finite subsets such that the generators involved in the relations in $R_0$ are all contained in $G_0$, ordered by inclusion. There is then a functor $F$ from $P$ to your category sending $(G_0,R_0)$ to an object with presentation $\langle G_0\mid R_0\rangle$, and it is easy to see that $X$ is a colimit of $F$.
Note also that any object is clearly the filtered colimit of its finitely generated subobjects. So if your algebraic theory further has the property that every finitely generated object is finitely presented, then any object is also the filtered colimit of its finitely presented subobjects. In particular, this is true of commutative rings (a finitely generated commutative ring is finitely presented because $\mathbb{Z}$ is Noetherian), or modules over any Noetherian ring. Since finitely generated commutative rings are also coherent, this also tells you any commutative ring is the colimit of its coherent subrings.
However, in general, an object need not be the colimit of its finitely presented subobjects. For instance, let $R=k[x_1,x_2,x_3,\dots]$ for some field $k$ and consider the $R$-module $M=R/(x_1,x_2,x_3,\dots)$. Then the only submodules of $M$ are $M$ and $0$, and $M$ is not finitely presented, so the colimit of its finitely presented submodules is just $0$.