You can always break up a long exact sequence into small exact sequences. Suppose your long exact sequence is:
$$\dots \to A_{n+1} \xrightarrow{f_{n+1}} A_n \xrightarrow{f_n} A_{n-1} \xrightarrow{f_{n-1}} A_{n-2} \to \dots,$$
then you can split it as:
$$0 \to \operatorname{coker} f_{n+1} \xrightarrow{f_n} A_{n-1} \xrightarrow{f_{n-1}} \operatorname{im} f_{n-1} \to 0.$$
Proving this is a bit tedious, but it follows directly from the definitions. So in some sense a long exact sequence lets you think of $A_{n-1}$ as an extension of the image of $f_{n-1}$ by the cokernel of $f_{n+1}$.
But as you can see that's not really insightful, and remembering the correct indices at all is a feat in itself (I had to revise what I wrote three times and I'm still not 100% sure). In this context long exact sequences are more of a computational tool than anything. As you say they're very useful, but giving them an interpretation like for short ones doesn't really work, in my opinion.
In homological algebra chain complexes are the basic objects of study, and a long exact sequence is a chain complex with zero homology; just like a space can have trivial homology but be very interesting, so can chain complexes...
I'd just like to point out an additional thing: there is something called the $\operatorname{Ext}$ functor that classifies extensions, that is if $A, B$ are modules over some ring $R$, then $\operatorname{Ext}^1_R(A,B)$ classifies extensions of the form $0 \to B \to X \to A \to 0$ (for some suitable notion of isomorphism of such extensions, see Wikipedia for details). More generally the $n$th $\operatorname{Ext}$ functor $\operatorname{Ext}^n_R(A,B)$ classifies extensions of the type:
$$0 \to B \to X_n \to \dots \to X_1 \to A \to 0$$
so in some sense you can think of bounded long exact sequences as higher extensions. (If you think about it, this is very related to the splitting I talked about at the beginning, that would allow you to do proofs by induction on the length of the sequence...)
Let's setup the framework. Let $\mathsf{Ab}$ be the category of abelian groups, and let $\mathsf{SES}$ be the category of short exact sequences, that is diagrams of the form $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ (with the usual conditions about images and kernels) and morphisms the obvious thing. Then define three functors,
$$\begin{align}
(H_n^a, H_n^b, H_n^c) & : \mathsf{SES} \to \mathsf{Ab}^3 \\
(A,B,C,f,g) & \mapsto (H_n(A), H_n(B), H_n(C))
\end{align}$$
The functoriality simply means that if you have three exact sequences and two chain maps $X \xrightarrow{\phi} X' \xrightarrow{\psi} X''$, then $H_n^x(\psi \circ \phi) = H_n^x(\psi) \circ H_n^x(\phi)$ for $x = a,b,c,$.
On the other hand the long exact sequence construction is a combination of three families of natural transformations, namely $f_* : H_n^a \to H_n^b$, $g_* : H_n^b \to H_n^c$ (both defined in the obvious way), and the connecting morphism $\partial_* : H_n^c \to H_{n-1}^a$. When one says that the long exact sequence is natural, one is really saying that these three families are all natural transformations.
Of course, you can set things up differently and construct a functor $\operatorname{LES} : \mathsf{SES} \to \mathsf{Ch}$ from the category of short exact sequences to the category of chain complexes, bundling everything together. But when people say that the long exact sequence construction is natural, they are implicitly thinking about all the $H_n$ separately.
This is in fact somewhat of a general construction. If you have two functors $F,G : \mathsf{C} \to \mathsf{D}$ and a natural transformation $\phi : F \to G$, then you can build up a new functor $\mathsf{C} \to \mathsf{Ar}(\mathsf{D})$ given by:
$$c \in \mathsf{C} \mapsto (F(c) \xrightarrow{\phi_c} G(c)) \in \mathsf{Ar}(\mathsf{D}),$$
and vice-versa (a functor $\mathsf{C} \to \mathsf{Ar}(\mathsf{D})$ is the same thing as two functors and a natural transformation).
It's hard to say which one is more "natural", but in general using this principle you can interpret any naturality result as a functoriality result, if you want – at the cost of using a more complicated target category.
Best Answer
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.