Suppose $\mathcal{E} \to \mathcal{C}$ is a cocartesian fibration over (the nerve of) a classical category, and there is a section on zero simplices that sends $C$ to $s(C)$ such that, for every edge $f\colon C \to C'$ in $\mathcal{C}$, the Kan complex


$\mathrm{Hom}\_{\mathcal{E}\_{C'}}(f\_!(s(C)), s(C'))$


is discrete (i.e. all higher $\pi\_i$ vanish).


Is it true that data of an extension $s$ to a full section $\mathcal{C} \to \mathcal{E}$ is equivalent to data of morphisms


$t\_f\colon [f\_!](s(C)) \to s(C')$


in the homotopy category $h\mathcal{E}\_{C'}$ such that, for every two composable morphisms $f\colon C \to C'$ and $g\colon C' \to C''$,


$t\_g \circ [g\_!](t\_f) ~= t\_{g\circ f}$


in the homotopy category $h\mathcal{E}\_{C''}$ (under the canonical identification of $[g\_!]\circ [f\_!]$ with $[(g\circ f)\_!]$).



Constructing sections of a cocartesian fibration
Yes. Consider the full subcategory $\mathcal S$ of $\mathcal E$ spanned by the objects $s(C)$.


It is clear that a section extending $s$ is the same thing as a section $\mathcal C\to \mathcal S$ extending $s$.


But I claim that $\mathcal S$ is a $1$-category: indeed the fiber of $\hom\_\mathcal E(s(C),s(C'))\to \hom\_\mathcal C(C,C')$ over $f$ is $\hom\_{\mathcal E\_C'}(f\_!s(C),s(C'))$, and so is discrete, and the base $\hom\_\mathcal C(C,C')$ is also discrete. Of course if $E\to B$ is a map to a discrete space where each fiber is discrete, $E$ must be discrete too (say by the long exact sequence of homotopy groups)


In particular, a section $\mathcal C\to \mathcal S$ extending $s$ is just an ordinary functor between ordinary categories : you have to pick images $s(f)$'s for every $f$ that compose well. Because $\mathcal S \subset \mathcal E$ and the latter is coCartesian over $\mathcal C$, you can reformulate the data of $s(f)$ in the way you did, and "compose well" accordingly.


