What are exact sequences, metaphysically speaking?

This was too long to put as a comment, I apologize if it doesn't help.

I don't know how totally accurate this is, but I like to think of (short) exact sequences as being algebraified versions of fiber bundles. Thus, putting $X$ in a short exact sequence $0\to Y\to X\to Z\to0$ indicates to me that $X$ is put together in some way from $Y$ and $Z$, and in such a way that, in a perfect world where everything is nice, is just the product of $Y$ and $Z$. Therefore, $X$ is some kind of "twisted product" of $Y$ and $Z$.

Thus, any time we are able to put $X$ into an exact sequene we should (in spirit) be able to tell properties of $X$ from properties of $Y$ and $Z$.

For example, knowing that $B$ is an abelian groups such that

$$0\to A\to B\to C\to 0$$

is a SES for $B,C$ also abelian groups tells me that $\text{rank}(B)=\text{rank}(A)+\text{rank}(C)$ (or more generally this works nicely for modules over PIDs).

The reason that SESs are such a convenient framework to deal with the notion of "put-togetheredness" is that we live in a fundamentally arrow obsessed world. Things phrased entirely in terms of arrows make us happy, because they are often easy to deal with.

One algebraic answer is that exact sequences are a natural abstraction of the notion of generators and relations. That is, let $R$ be a ring and $M$ a left $R$-module with generating set $S$. Then there is a canonical surjection $$R^S \xrightarrow{f} M \to 0.$$

The kernel of this surjection describes all the possible relations in $S$ and gives rise to a short exact sequence $$0 \to \text{ker}(f) \to R^S \xrightarrow{f} M \to 0.$$

If $R$ is a PID, then $\text{ker}(f)$ is free, so picking a basis for $\text{ker}(f)$ gives an irredundant set of relations among the generators. However, if $\text{ker}(f)$ is not free, then picking a defining set of relations $T$ (that is, a generating set in $\text{ker}(f)$) instead gives rise to an exact sequence $$0 \to \text{ker}(g) \to R^T \xrightarrow{g} R^S \xrightarrow{f} M \to 0.$$

If $\text{ker}(g)$ is not free, then... and so on. From this perspective we are thinking of exact sequences as resolutions.