Let $\mathcal{C}$ be an $\infty$-category, and let $G$ be a group. Denote by $\mathcal{C}^{\text{BG}}$ the functor $\infty$-category $\text{Fun}(\text{BG}\longrightarrow \mathcal{C})$. The homotopy orbit functor is given by
$$-_{\text{hG}}: \mathcal{C}^{\text{BG}}\longrightarrow \mathcal{C}: (F: \text{BG}\longrightarrow \mathcal{C})\mapsto \text{colim}_{\text{BG}}F.$$
What does taking a $\text{colim}$ with respect to $\text{BG}$ mean? What type of object does it represent, and what is the relation of this functor to the classical concept of orbits of groups acting on a set or a space?
The object $BG$ is a space, and hence an $\infty$-groupoid, so we can index diagrams using it. One particular presentation of this homotopy type is as the Kan complex $N(\mathbf{B}G)$, where $N(-)$ is the nerve functor and $\mathbf{B}G$ is the $1$-object category with $\mathrm{Hom}_{\mathbf{B}G}(*,*)=G$. This means that $N(\mathbf{B}G)_n\cong G^{\times n}$ for all $n\geq 0$. The simplicial structure maps are the usual ones that a nerve of a category has. So a functor $F\colon BG\to\mathcal{C}$ can be thought of as picking a single object $x\in\mathcal{C}$, picking for every $g\in G$ an automorphism $\theta_g\colon x\xrightarrow{\sim}x$, for every $f,g\in G$ a homotopy $\theta_g\theta_f\simeq\theta_{gf}$, for every $f,g,h\in G$ a further homotopy witnessing our already defined homotopies $\theta_h\theta_g\theta_f\simeq\theta_{hg}\theta_f\simeq\theta_{hgf}$ and $\theta_h\theta_g\theta_f\simeq\theta_h\theta_{gf}\simeq\theta_{hgf}$ are actually themselves homotopic, etc. In this sense, a map $BG\to\mathcal{C}$ picks out an object $x\in\mathcal{C}$ and equips it with a fully homotopy coherent $G$-action. (In my notation, it would be a left $G$-action.)
A cocone under such a functor $BG\to\mathcal{C}$ which equips $x\in\mathcal{C}$ with such a $G$-action is an object $y\in\mathcal{C}$ together with a map $p\colon x\to y$ in $\mathcal{C}$, and the following extra data: for every $g\in G$ we ask for a homotopy $p\theta_g\simeq p$, and for every $g,f\in G$ we ask for a witness that the two homotopies $p\theta_g\theta_f\simeq p\theta_f\simeq p$ and $p\theta_g\theta_f\simeq p\theta_{gf}\simeq p$ are themselves homotopic (here we used the homotopy $\theta_g\theta_f\simeq\theta_{gf}$ that is part of the functor $BG\to\mathcal{C}$). Similarly, for $h,g,f\in G$, we have a similar coherence requirement, and we keep doing asking for those for arbitrarily many finite lists of elements of $G$.
The colimit of our functor $BG\to\mathcal{C}$ is the initial of such data $(y,p)$. Informally, as Thorgott pointed out above, this means that the colimit of $BG\to\mathcal{C}$ just quotients out the orbits of $x$, but it does so in a homotopical way. Informally, if $a,b\colon *\to x$ are points of $x$ and $g\in G$ is such that $\theta_g\circ a\simeq b$, then $x_{hG}$ will have corresponding points $[a],[b]\colon *\to x_{hG}$ and the data of a path between them that corresponds to the witness of $\theta_g\circ a\simeq b$.
For instance, if we take the terminal object $*\in\mathsf{Spc}$ in the $\infty$-category of spaces and equip it with a trivial $G$-action, then $*_{hG}\simeq BG$! So our homotopy orbits have much more structure than the strict orbits, as $BG$ is generally a very large object, while the strict orbits would just give us a single point $*$. Intuitively the reason that $*_{hG}\simeq BG$ is that $*_{hG}$ will have a point for every point of $*$ (so it has a single point), and for every $g\in G$ it will add a path from this point to itself to witness that, in the original space $*$, the element $g$ maps this point to itself. Hence the $1$-morphisms of $*_{hG}$ are the elements of $G$. Likewise, you can show that its $2$-simplices (presenting a space as a Kan complex) are given by elements of $G\times G$, etc. Of course, if you formally want to argue that $*_{hG}\simeq BG$, we just note that $*_{hG}\simeq\mathrm{colim}_{BG} \,*\simeq BG$, as for any space $X$ it holds that $\mathrm{colim}_X\,*\simeq X$ in $\mathsf{Spc}$. This is for instance a consequence of the straightening-unstraightening equivalence.
If $X$ is a topological space with $G$-action, then the strict orbit space $X/G$ is a cocone under $BG\to\mathsf{Spc}$ in the $\infty$-category $\mathsf{Spc}$, and hence we get a canonical map $X_{hG}\to X/G$ in $\mathsf{Spc}$. This map is, as the example above shows, generally far from being an equivalence, and the source is generally much larger than the target (and contains more interesting information). If $X$ is merely a set with $G$-action, then we can recover the set $X/G$ as $\pi_0(X_{hG})$, where we consider $X$ as a discrete space to have it live in $\mathsf{Spc}$ (and we take the homotopy orbit there). This follows from the fact that $\pi_0\colon \mathsf{Spc}\to\mathsf{Set}$ is a left adjoint functor, and hence commutes with colimits. For a general $\infty$-category $\mathscr{I}$, a functor $\mathscr{I}\to\mathsf{Set}$ necessarily factors as a composite $\mathscr{I}\to\mathrm{Ho}\,\mathscr{I}\to\mathsf{Set}$, where $\mathrm{Ho}\,\mathscr{I}$ is the homotopy $1$-category of $\mathscr{I}$. This is because $\mathsf{Set}$ is itself a $1$-category. Hence our functor $BG\to\mathsf{Spc}\xrightarrow{\pi_0}\mathsf{Set}$ factors through a functor $\mathrm{Ho}\,BG\to\mathsf{Set}$. But $\mathrm{Ho}\,BG\simeq \mathbf{B}G$ as $1$-categories, as can for instance seen via the presentation $BG\simeq N(\mathbf{B}G)$ from the first paragraph. Hence $\pi_0(X_{hG})$ is isomorphic in $\mathsf{Set}$ to $\mathrm{colim}_{\mathbf{B}G} X$ (where the $X$ on the right hand side is the good old set $X$ that we started with). The latter object is just $X/G$ as Thorgott explained in his comment, so this finishes the proof of our claim. So in case of $G$-actions on sets (with $G$ a discrete group), the homotopy orbits are a natural way to turn the set $X/G$ into a space with richer structure, whose path components give us back the original orbits of the $G$-action.