I'll write $y$ instead of $s$ and $x$ instead of $t$. You wonder, when can the solution of
$$\tag{1}
y^{(n)}(x)=f(y)
$$
be reduced to quadratures? (As you have done for $n=1,\ 2$ in OP). For $n=2$ you used a conserved quantity (energy) to reduce the order of the equation. The 'general (theoretical) procedure' is to find as many conserved quantities as possible, and use each to reduce the order by one. For $n\geq 3$ only one conserved quantity is known, and that's only when $n$ is even- so the short answer is: there is no complete general procedure of the type you describe for $n\geq 3$.
Looking in Polyanin and Zaitsev: Handbook of exact solutions to ODEs eq. $5.2.6.8$ we find that
$$\tag{2}
C=\int dy \ f(y)+\sum\limits_{k=1}^{m-1}(-1)^ky^{(k)}y^{(2m-k)}+\frac{(-1)^m}{2}\left[y^{(m)}\right]^2
$$
is a first integral of (1) when $n=2m$. Equation (2) is an autonomous ODE of order $n-1$, which can be reduced to a non-autonomous order $n-2$ ODE by the usual substitution $y'(x)=v(y)$.
Another 'trick' is to recast (1) as $n$ coupled first order equations. Define $y^{(k)}=y_k$ for $k=0,1, \dots n-1$, then we have the system
$$\tag{3}
y_0'=y_1 \\
y_1'=y_2 \\
\cdots \\
y_{n-1}'=f(y_0)
$$
A formal solution to (3) may be written using an exponential operator
$$\tag{4}
y_k(x)=\exp\left(x y_1 \frac{\partial}{\partial y_0} +x y_2 \frac{\partial}{\partial y_1}+\cdots +xf(y_0)\frac{\partial}{\partial y_{n-1}}\right) \ y_k(0)
$$
where the exponential of an operator $A$ is defined using the power series
$$\tag{5}
\exp(A) =\sum\limits_{j=0}^\infty \frac{A^j}{j!}
$$
See chapter 5, CJ Papachristou: Aspects of integrability for details. By truncating the exponential power series, eq. (4) furnishes a systematic method of generating approximate solutions to (3) in powers of $x$ (ie. as $x\to 0$).
A concise discussion of (1), its Lagrangian and Hamiltonian, and the problem with $n\geq 3$ may be found in this article by G Lopez.
