Object of a Category $C$ acts as Functor

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I have a question about a notation used in following paper: https://etale.site/writing/stax-seminar-talk.pdf (see page 4):

We take a category $C$ and consider a pair $(X_0,X_1)$ of two objects $X_1, X_1 \in C$ satisfying identities between maps $s, t , \epsilon, i $ and $m$ as described in the article above.

Then we define a new category in following way:

We take a $U \in C$ and define a category $\{X_0(U)/X_1(U)\}$ such that:

The objects are defined via $ob(\{X_0(U)/X_1(U)\}):= X_0(U)$.

And exactly this is the problem: What is exactly $X_0(U)$? Especially how $X_0$ "acts" on $U \in C$?

Is $X_0$ concretely an element in $C$ or a functor $X_0:C \to (Set)$, $U \mapsto X_0(U)$?

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To expand on my comment, $X_0(U)=\newcommand\Hom{\operatorname{Hom}}\Hom(U,X_0)$. That is, we identify objects of a category with their functors of points.

To rephrase the definition here, $X_0$ and $X_1$ are objects in $C$ parametrizing objects and morphisms of the groupoid respectively. Applying the functor $\Hom(U,-)$ to our groupoid, we see that for all $U$, $X_0(U)$ and $X_1(U)$ together with $s,t,\epsilon,i,m$ determine a groupoid in $\mathbf{Set}$. This is the category $\{X_0(U)/X_1(U)\}$. Since $\Hom(U,-)$ is also functorial in $U$, for morphisms $f:V\to U$, we get pullback maps $f^* : \{X_0(U)/X_1(U)\}\to \{X_0(V)/X_1(V)\}$.