Definition of $E_n$-modules for an $E_n$-algebra

$E_n$ algebras have compatible multiplications for every way of placing a bunch of elements into a collection of balls in $\mathbb{R}^n$. A module for an $E_n$ algebra has an action for every way of placing a bunch of elements of the algebra into balls not at the origin, and an element of the module into a ball at the origin.

Here's what this looks like in a few examples to help you gain some intuition.

• $E_1$ algebras in vector spaces are ordinary (non-commutative) algebras. An $E_1$-module for $A$ is just an $A$-$A$ bimodule. I.e. there's actions for every way of placing another ball not at the origin (so either to the left or right) and you get an associativity axiom for every way of placing two balls not at the origin (both left is the left module axiom, both right is the right module axiom, and one on each side is the bimodule axiom).
• $E_2$ algebras in vector spaces are commutative algebras (because you can continuously deform two discs in the plane into the opposite order, so $xy = yx$). An $E_2$-module for $A$ is just an $A$-module (which is also automatically a right $A$-module by using commutativity to turn the left action into a right action).
• $E_1$ algebras in categories are monoidal categories. $E_1$-modules for monoidal categories are bimodule categories in the sense of Chapter 7 of tensor categories. I.e. you have bifunctors $\vartriangleright: A \times M \rightarrow M$ and $\vartriangleleft: M \times A \rightarrow M$ together with natural associators for each triple product (e.g. $\alpha_{a,b,m}: a \vartriangleright (b \vartriangleright m) \rightarrow (a \otimes b) \vartriangleright m)$ satsifying some hexagon axioms for quadruple products.
• $E_2$ algebras in categories are braided monoidal categories. Both horizontal and vertical composition are the monoidal structure, but the braiding tells you how 180-degree rotation gives a braiding iso $\beta_{x,y}: x \otimes y \rightarrow y \otimes x$ satisfying a hexagon axiom for triple tensor products. Note that you can either rotate clockwise or counterclockwise, yielding the braiding and its inverse, but these need not agree. Now let's think about $E_2$ modules. The action comes from a pair of discs, one of which is a marked disc at the origin. I.e. we want a left module category $M$. The important isotopy here is no longer the swap, but instead a full rotation around the marked disk (i.e. the generator of the affine braid group on one strand). So we need a full-twist isomorphism $\eta_{a,m}: a \vartriangleright m \rightarrow a \vartriangleright m$. But looking at three discs, this needs to satisfy a compatibility axiom related to the defining relation of the 2-strand affine braid group which states that $\eta_{a\otimes b,m}: (a \otimes b) \vartriangleright m \rightarrow (a \otimes b) \vartriangleright m$ is equal to the composite: $$(a \otimes b) \vartriangleright m \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow a \vartriangleright (b \vartriangleright m) \rightarrow (a \otimes b) \vartriangleright m \rightarrow (b \otimes a) \vartriangleright m \rightarrow (a \otimes b) \vartriangleright m$$ where the first and fourth maps are associators and the second and third maps are the full-twists $\mathrm{id}_a \vartriangleright \eta_{b,m}$ and $\eta_{a, b\vartriangleright m}$ (or possibly vice-versa), and the fifth and sixth maps are both the braiding (or possibly both the inverse braiding). This structure is called a "braided module category" by Enriquez (see also Brochier) and the translation between $E_2$-modules and braided module categories is given in Theorem 3.11 of Ben-Zvi-Brochier-Jordan.
• $E_3$ algebras in categories are symmetric monoidal categories. I.e. braided tensor categories where the braiding and the inverse braiding agree. Similarly, for $E_3$-modules the full-twist maps are automatically trivial by untying them in the extra dimension. So an $E_3$-module for a symmetric tensor category is just a module category.

(Note that in vector spaces $E_2$-algebras are $E_\infty$-algebras, while in categories $E_3$-algebras are $E_\infty$-algebras. So in both cases we see that $E_\infty$-modules for $E_\infty$-algebras agree with the usual notion of left-module = right-module, but below the stable range the notion is subtler and is a bimodule with the appropriate amount of compatibility between the left and right module structures.)

$\newcommand{\E}{\mathbf{E}} \newcommand{\Mod}{\mathrm{Mod}} \newcommand{\cc}{\mathcal{C}}$Here's one way to think about $\E_n$-modules. Let $R$ be an $\E_n$-ring (in a presentable symmetric monoidal category $\cc$), and let $\Mod^{\E_n}_R(\cc)$ denote the category of $\E_n$-$R$-modules. There is a forgetful functor $\Mod^{\E_n}_R(\cc) \to \cc$, whose left adjoint is the free $\E_n$-$R$-module; then, $\Mod^{\E_n}_R(\cc)$ is equivalent to $\mathrm{L}\Mod_{\mathrm{Free}^{\E_n}_R(1_\cc)}(\cc)$. In the case $\cc = \mathrm{Sp}$, one way to see this is as follows: the image of the unit in $\cc$ (i.e., the sphere spectrum $\mathbb{S}$) is a compact generator of $\Mod^{\E_n}_R(\mathrm{Sp})$, so that category is equivalent to the category of left modules over the endomorphism ring of $\mathrm{Free}^{\E_n}_R(\mathbb{S})$ in $\Mod^{\E_n}_R(\mathrm{Sp})$. But this endomorphism ring is just $\mathrm{Free}^{\E_n}_R(\mathbb{S})$.

Returning to the general case, we're reduced to understanding $\mathrm{Free}^{\E_n}_R(1_\cc)$. This object is what's known as the "enveloping algebra"; it's denoted by $U_R$ in Section 3 of this paper. There, it's shown that $\mathrm{Free}^{\E_n}_R(1_\cc) \simeq \int_{S^{n-1}} R$ (factorization homology). The intuitive interpretation of this identification is that an $\E_n$-$R$-module (i.e., a left $\mathrm{Free}^{\E_n}_R(1_\cc)$-module) is an object of $\cc$ with a collection of commuting left actions of $R$, where these actions are parametrized by $S^{n-1}$.

When $n=1$, this factorization homology is easy to describe. Indeed, $S^{n-1} = S^0$ is just a disjoint union of two points, and then $\mathrm{Free}^{\E_1}_R(1_\cc) \simeq \int_{S^0} R = R\otimes R^{op}$. In other words, $\E_1$-$R$-modules (for an $\E_1$-algebra $R$ in $\cc$) are the same thing as left $R\otimes R^{op}$-modules, i.e., $R$-$R$-bimodules in $\cc$.

If you accept an intuition that comes from plain (non-$\infty$) operads, then this is relatively easy. For an $E_n$-algebra $A$, an $E_n$-module over $A$ is an object $M$ equipped with structure maps $E_n(k+1+l) \otimes A^{\otimes k} \otimes M \otimes A^{\otimes l} \to M$ that satisfy relatively obvious associativity and equivariance conditions if you draw pictures with trees.

For example with $n=1$, you have a configuration of intervals in $[0,1]$. In one of the intervals, you can plug an element of $M$, and in the others you plug elements of $A$. Up to homotopy, all that matters is the order of the intervals, so you get maps $A^{\otimes k} \otimes M \otimes A^{\otimes l} \to M$ for $k,l \ge 0$. The $A$'s on the left gets you a (homotopy) left $A$-module structure, and the $A$'s on the right get you a (homotopy) right $A$-module structure. The axioms for an $E_1$-module tell you that these actions are compatible with the $E_1$-algebra structure of $A$. If $A$ is an ordinary algebra, then these axioms are telling you that $(a \cdot a') \cdot m \simeq a \cdot (a' \cdot m)$, the same thing on the other side, and something like $(a \cdot m) \cdot a' \simeq a \cdot (m \cdot a')$, for example. If all these higher operations are trivial, then you really get an $(A,A)$-bimodule structure.