For me, it's fairly simple, really. Suppose you have the problem
$$
L u = a(x) u''(x) + b(x) u'(x) + c(x) u = f(x),
$$
with boundary conditions
$$
B u = \begin{pmatrix} \alpha_{11} u(x_1) + \alpha_{12} u'(x_1) \\ \alpha_{21} u(x_2) + \alpha_{22} u'(x_2) \end{pmatrix} = 0,
$$
where $a$, $b$, $c$ and $f$ are smooth enough, in order for all the following calculations to make sense, and $\alpha_{11}\alpha_{22} - \alpha_{12}\alpha_{21} \neq 0$. In a very broad sense, we are thinking of an operator $L$ that takes a function $u$ in the space $X = \{ u \text{ smooth enough such that } B u = 0 \}$ and send it to the function $f(x)$ in a space $Y$ (with the needed characteristics, I'm not going to get technical here):
$$
L: X \longrightarrow Y.
$$
In this setting, a fair question is "What are the properties of $L$?"
If we define the inner product as
$$
\langle u(x), v(x) \rangle := \int_{x_1}^{x_2} u(x)v(x) dx
$$
and the adjoint $L^*$ by the right hand side of the equation
$$
\langle u(x), L^* v(x) \rangle := \langle L u(x), v(x) \rangle,
$$
then, with the proper definitions, we can study $L$ in the same way we studied matrices. So, let's take a look to the expression $\langle L u(x), v(x) \rangle$. Expanding, we have
\begin{multline}
\langle L u(x), v(x) \rangle = \big(a u'v - u (a v)' + u(bv)\big)\big|_{x_1}^{x_2} \\
+ \int_{x_1}^{x_2}\big(a v'' - (2 a' - b)v' + (a'' - b' + c)v\big)u dx
\end{multline}
Then,
$$
L^* v = a v'' + (2 a' - b)v' + (a'' - b' + c)v
$$
and $B = B^*$ (can you prove this?).
As in matrices, a very special kind of operator is the self-adjoint operator, $L = L^*$. Why? Well, among other things, because all its eignevalues are real and all its eigenfunctions are orthogonal (you should prove this also). Now, what does the self-adjoint operator looks like for second order ODEs? The function $a$ and $b$ must satisfy
\begin{align}
2 a' - b &= b,\\
a'' - b' + c &= c.
\end{align}
Thus, $a' = b$, and $L$ can be written as
$$
L u = a u'' + a' u' + c u = (a u')' + c u,
$$
which is a Sturm-Liouville operator!
So, Sturm-Liouville operators are self-adjoint operators and have all the fantastic properties of self-adjoint operators. Great! But, they seem rather rare, as the condition $a' = b$ seems pretty restrictive. Why on Earth, then, we take so much effort in studying them? Well, what if instead of the regular inner product that I took, I use the inner product
$$
\langle u(x), v(x) \rangle_{w(x)} = \int_{x_1}^{x_2} u(x) v(x) w(x) dx,
$$
where $w(x) > 0$?
Then,
$$
w L u = w a u'' + w b u ' + w c u = (w a u')' + (w b - (w a)')u' + w c,
$$
and if $w$ satisfy the ODE
$$
a w' + (b - a') w = 0,
$$
the operator $wL$ is selfadjoint!
In conclusion,
With the proper inner product ($w(x) = e^{\int \frac{b-a'}{a} dx}$), every second order ODE can be studied as a Sturm-Liouville problem, where the operator is self-adjoint!
With one stroke, we've enclosed all second order ODEs into one, which happens to be self-adjoint. Those guys really knew what they were doing!