Suppose $\mathcal{D}$ is an $\infty$-category with finite limits. Given a pointed object $\ast\rightarrow A$, its Cech nerve is a simplicial object $M_\bullet$. By Higher Topos Theory, Proposition 6.1.2.11, it is actually a groupoid and, moreover, since $M_0\cong \ast$, it is a group. This gives $M_1\cong \Omega A$ the structure of a group.
Given a morphism $B\rightarrow A$ consider the pullback diagram
$$
\require{AMScd}
\begin{CD}
B\times_A \ast @>>> B\\
@VVV @VVV \\
\ast @>>> A
\end{CD}
$$
Taking the Cech nerves horizontally you get a morphism of simplicial objects $M'_\bullet\rightarrow M_\bullet$. Using that $M'_\bullet$ is a groupoid, it is easy to see that $M'_\bullet\rightarrow M_\bullet$ is a left action object in the sense of Higher Algebra, Definition 4.2.2.2. So, the group $M_1\cong \Omega A$ acts on $M'_0\cong B\times_A \ast$.
Let's restrict to the case $\mathcal{D} = \mathcal{C}_{/X}$. Consider the object $A=(p_1\colon X\times X\rightarrow X)$. The final object in $\mathcal{C}_{/X}$ is $\mathrm{id}\colon X\rightarrow X$ and $A$ has a pointing by the diagonal map $\Delta\colon X\rightarrow X\times X$. Its based loop space is $\Omega_\Delta A = (\mathcal{L} X\rightarrow X)$.
Note that using that $\mathcal{C}$ is cotensored over spaces, you can obtain the Cech nerve of $X\rightarrow X\times X$ in $\mathcal{C}_{/X}$ by applying $\mathrm{Map}(-, X)$ to the Cech conerve of $S^0\rightarrow \ast$ in pointed spaces which gives $\Sigma S^0\cong S^1$ the structure of a cogroup in pointed spaces.
Given a morphism $f\colon Y\rightarrow X$ in $\mathcal{C}$, let $B = (p_1\colon X\times Y\rightarrow X)$ with $B\rightarrow A$ given by $(\mathrm{id}\times f)\colon X\times Y\rightarrow X\times X$. Then $B\times_A \ast = (X\times Y)\times_{X\times X} X\cong Y$ which by above gives an action of $\mathcal{L} X\rightarrow X$ on $Y\rightarrow X$.
Note that you could instead consider $\tilde{B} = (p_1\circ(f\times \mathrm{id})\colon Y\times X\rightarrow X)$. This will give the trivial action of $\mathcal{L} X\rightarrow X$ on $Y\rightarrow X$.
In algebraic geometry you can think of it as follows. Let's say $X$ is a (derived) scheme. $\widehat{X\times X} = [X/\mathcal{L} X]$ is the classifying space of the group scheme $\mathcal{L}X\rightarrow X$ and $\widehat{X\times Y} = [Y/\mathcal{L} X]$.
