Let me focus on the sentence:
Is it possible to get something like a concrete example of how the
higher order morphisms that make up the operad determine how the lower
order ones behave?
Let us have a look at operations on a chain complex $C$. Such an $n$-ary operation is just an elemeent of the Hom-complex $\operatorname{Hom}(C^{\otimes n},C)$ and composition is defined in the canonical way. To avoid sign issues I want to work with $\mathbb{F}_2$-coefficients.
I would like to argue now that the surjection operad is such an interesting example.
If $C^*$ happens to be the normalized cochain complex of a simplicial set $X$, we can write down a look of such interesting operations. The first example is the cup-product, defined as
$$\cup: C^m\otimes C^n \to C^{m+n}\qquad (\varphi\cup \psi)(\sigma:\Delta^{m+n}\to X) := \varphi(\sigma|_{0,\ldots,m})\cdot \psi(\sigma|_{m,\ldots,m+n}).$$
This is clearly an interesting example, as it induces a multiplication on homology.
The surjection operad now generates a lot of such operations in a similar manner. Let us have a look at their definitions. First we use the convention that a cochain evaluated on a chain of different dimension returns zero.
Now fix a surjection $s:\{1,\ldots,m+k\}\to \{1,\ldots,m\}$, such as $121$. This is an overly abbreviated notation for the map sending $1$ to $1$, $2$ to $2$ and $3$ to $1$. As long as $m<10$ this notation still works and we won't use such high $m$'s in this example anyway.
Associated to this surjection we would like to define a operation $\langle s\rangle \in \operatorname{Hom^{-k}}(C^{\otimes m},C)$. So we have to define what
$\langle s\rangle(\varphi_1\otimes\ldots\otimes \varphi_m)(\sigma:\Delta^N\to X)$ is.
Think of $0,\ldots,N$ as $N$ succesive edges and look at all the ways to cut them into $m+k$ parts. For the example of 121, we thus have to pick two numbers $0\le n_1\le n_2\le N$.
Then use the given surjection to decide which part gets fed into which of the $\varphi_i$'s. For $121$ we would get:
$\langle 121\rangle(\varphi_1\otimes\varphi_2)(\sigma:\Delta^n\to X) = \sum_{0\le n_1\le n_2\le N}\varphi_1(\sigma|_{0,\ldots,n_1,n_2,\ldots,N})\varphi_2(\sigma|_{n_1,\ldots,n_2})$
with the convention that if $n_1=n_2$, then the argument of $\varphi_1$ is a degenerate simplex and thus the normalized cochain vanishes on it (normalized means it vanishes on all degenerate simplices).
Question/Exercise: For which $s$ is $\cup=\langle s\rangle$?
So why are these operations useful?
There is a formula to express $d\langle s\rangle\in \operatorname{Hom}^{-k+1}(C^{\otimes m},C)$ as a linear combination of operations $\langle s'\rangle$ for $s'\{1,\ldots,m+k-1\}\to\{1,\ldots,m\}$, namely you sum over all elements of $1,\ldots,m+k$, remove that given element, and check if the given map is still surjective. Then you add up all these choices.
So for example $d(\langle 121\rangle) = \langle 12\rangle+\langle 21\rangle$. Translating back what this means on cochains is that
$d(\langle 121\rangle (\varphi_1\otimes \varphi_2)) = \varphi_1\cup \varphi_2 +\varphi_2\cup \varphi_1$, e.g. while the multiplication on $C$ is not commutative, the difference is always a boundary element. So the operation $\langle 121\rangle$ determines how $\langle 12\rangle$ behaves.
Indeed $\langle 121\rangle$ is also called $\cup_1$ and $\langle 1212\rangle=\cup_2$ and so on. And so these operations can also be used to define Steenrod-operations on $C^*$. There are relations between the Steenrod-operations on homology, like the Adem- and Cartan-relations. They typically do not hold on chain-level but only up to a boundary of some element $x$. Often to such an relation one can consider a higher operation. To explicitly compute that higher operation, one needs a way to construct that element $x$. This can often be done using the surjection operad.
In some sense this is an example of how the higher order morphisms determine how the lower order ones behave. I just changed the definition of higher from more arguments to higher in as in the degree in the Hom-complex.
If you also want higher to go in the direction of more arguments, then one really has to look at the composition formulas.
