Is there some practical intuition when working with a cooperad given by cogenerators and corelations? In the case of algebras and operads, a description by generators and relations is common practice and I have a good understanding of this. A non-symmetric operad $\mathcal{P}$ given by a linear space of generators $E$ and relations $R \subset \mathcal{T}(E)$ is defined as the quotientoperad $\mathcal{T}(E)/(R)$ where $(R)$ is the operadic ideal generated by $R$.
The intuition behind this construction is that elements of $\mathcal{P}$ are operadic compositions of generators where the relations "hold", that is, are set to $0$.
Dually, I have great difficulty understanding a non-symmetric cooperad $\mathcal{C}$ cogenerated by a linear subspace $E$ and corelations $R \subset \mathcal{T}^c(E)$, where $\mathcal{T}^c$ is the free non-symmetric cooperad functor. I even seem to have difficulty with stating the formal definition of $\mathcal{C}$.
In Algebraic Operads (Loday-Vallette), this construction is used for defining the Koszul dual cooperad (7.1.4). In that case, we have $R \subset \mathcal{T}^c(E)^{(2)}$ and $\mathcal{C}$ is the cooperadic kernel of the map $\mathcal{T}^c(E) \twoheadrightarrow \mathcal{T}^c(E)^{(2)}/ R$. However, is this map even a morphism of cooperads? And what if $R$ is not concentrated in weight $2$?
If somebody can provide a nice framework with intuiton for this construction or a concrete reference, that would help me a great deal!
Edit: an example of practical intuition: If $E$ is generated (as linear space) by $x$ and $y$ (in some arity), and $R= F(x,y)$ some 'operadic polynomial' in $x$ and $y$, then we know that $\mathcal{P}$ is generated by two elements $X,Y \in \mathcal{P}$ and composing $X$ and $Y$ accoding to the  polynomial $F(x,y)$ will equal $0$ in $\mathcal{P}$.
Analogously, if again $E$ is generated by $x$ and $y$ and $R= F(x,y)$, then we know that $x,y\in \mathcal{C}$ and $F(x,y)\in \mathcal{C} \hookrightarrow \mathcal{T}^c(x,y)$. However, do we always then have that $$\overline{\Delta}^\mathcal{C}(F(x,y))= F(x,y) \in \mathcal{T}^c(\mathcal{C})$$
where $\overline{\Delta}^\mathcal{C}$ is the reduced operadic decomposition of $\mathcal{C}$? For example for the Koszul dual cooperad $Assoc^c$, it is cogenerated by an element $m^c$ of arity $2$ with corelations $m^c_3 := m^c \circ_1 m^c - m^c \circ_2 m^c$ in arity $3$. Moreover,  we have $\overline{\Delta}(m^c_3) =  m^c \circ_1 m^c - m^c \circ_2 m^c $.
 A: Cooperads are a bit more tricky than operads, but they are useful.
First, notice that if $E$ is any collection and $R$ is a subcollection of the free conilpotent cooperad $\mathcal T_E^c$ (which is equal to the free operad $\mathcal T_E$ as a collection), then the conilpotent cooperad cogenerated by $E$ subject to relations $R$ is a subcooperad of $\mathcal T_E^c$. Dually, the operad generated by $E$ subject to $R$ is a quotient of $\mathcal T_E$.
The question, then, is exactly which subcooperad is it? First, notice that even in case $R$ is quadratic, it is definitely not the kernel of the map
$$\mathcal{T}^c_E \twoheadrightarrow \mathcal{T}^{c,(2)}_E/ R.$$
The right hand side is not a cooperad, and the kernel is too big in any case (do it in the non-symmetric case for the associative operad, for example). Although this is rather informal, once you fix $E$ and $R$, you have that
$$\mathcal C^{(1)} = E, \quad \mathcal C^{(2)} = R$$
and just like how in the case of operads you want to quotient out by the largest thing such that the composition kills every tree where $R$ appears, so its kernel is big but not too big, in the case of conilpotent cooperads, you want to only keep the smallest thing that guarantees that the decomposition does not fall out of $\mathcal C$, so its image is small, but not too small.
To explain how the construction is rather complicated, consider $\mathsf{Ass}$, as you did. Then $\mathsf{Ass}^c$ is one dimensional in each arity, and in fact you can show that there exists a choice of signs so that the unique generator $\mu_n^c$ in arity $n$ is of the form
$$\sum_{T\in \mathsf{PRBT}_n} (-1)^{\varepsilon(T)} \mu_T$$
where $T$ runs through planar rooted binary trees with $n$ leaves. For example, you correctly stated that
$$\mu_3^c = (x_1x_2)x_3 - x_1(x_2x_3),$$
and that its reduced decomposition is equal to $\mu_2^c\circ_1 \mu_2^c-
\mu_2^c \circ_2\mu_2^c$.
With this choice of generator, I think that
$$\mu_4^c = ((x_1x_2)x_3)x_4 - (x_1(x_2x_3))x_4  +
  x_1((x_2x_3)x_4) - x_1(x_2(x_3x_4)) - (x_1x_2)(x_3x_4)$$
works. If you compute the decomposition map of $\mu_4^c$ you will just get the usual formula for Stasheff's $A_\infty$-relation involving $\mu_2^c$ and $\mu_3^c$ and five terms.
The best way to think about $\mathcal C$ as a kernel in the weight graded case is to take $\mathcal P = \mathcal T_E/(R)$, form the bar construction $\mathsf{B}(\mathcal P)$ and give it the syzygy grading, and then prove that $\mathcal C$ is isomorphic to the zeroth cohomology group $H^0(\mathsf{B}(\mathcal P)) = \ker\partial^0$.
From this description is the clear that $\mathcal C$ is equal to $E$ in weight $1$ and to $R$ in weight $2$, and that one must take summands that conspire to create relations once you decompose them, but the way one does this is a bit more tricky, and the description
$$\mathcal C^{(n)} = \bigcap_{j=1}^{n-2} 
V^{\otimes j}\cap R\cap V^{n-j-2}$$ of the conilpotent coassociative coalgebra generated by $V$ subject to $R$ as the intersection of the individual summands of the face maps is no longer available.
A good way to explicitly write down dual cooperads is to simply observe that as collections they are dual to the dual operad, and that when things are locally finite dimensional, one can simply write the decomposition map as the dual of the composition map, by simply taking some basis operation $\eta$ and finding all ways in which it can be written as the composition $\gamma(\eta_0;\eta_1,\ldots,\eta_k)$ of other basis operations.
In this way, for example, one can explicitly write down the decomposition maps of operads like $\mathsf{Lie},
\mathsf{Com}$ and $\mathsf{Poiss}$  without having to explicitly write down a $0$th cohomology representative, which usually involves a sum of many terms, as you see above.
