A functor is a fibration if every morphism in the base to the image of an object in the total category lies below a cartesian morphism to the object in the total category.
Thus, it is enough to understand the cartesian morphisms of the functor.
Recall that given a functor $\mathcal E\xrightarrow{p}\mathcal C$, a morphism $Y\xrightarrow{\alpha} Z$ in $\mathcal E$ is $p$-cartesian if whenever $X\xrightarrow{\gamma}Z$ in $\mathcal E$ is such that $p(X)\xrightarrow{p(\gamma)}p(Z)$ in $\mathcal C$ factors as $p(X)\xrightarrow{b}B\xrightarrow{p(\alpha)}p(Z)$ in $\mathcal C$, there is a unique morphism $X\xrightarrow{\beta}Y$ in $\mathcal E$ with $p(\beta)=b$ and $X\xrightarrow{\gamma}Z$ factoring as $X\xrightarrow{\beta}Y\xrightarrow{\alpha}Z$.
In your case, a cartesian morphism above a functor $J\xrightarrow{a}K$ given a functor $K\xrightarrow{Z}C$ will be a natural transformation $Y\overset{\alpha}\Rightarrow Z\circ a$ for functors $\mathcal C\xleftarrow{Y}\mathcal J\xrightarrow{a}\mathcal K\xrightarrow{Z}\mathcal C$ so that for each $X\overset\gamma\Rightarrow Z\circ a\circ b$ for functors $\mathcal C\xleftarrow{X}\mathcal I\xrightarrow{b}\mathcal J\xrightarrow{a}\mathcal K\xrightarrow{Z}\mathcal C$, there is a unique $X\overset\beta\Rightarrow Y\circ b$ for $\mathcal C\xleftarrow{X}I\xrightarrow{b}\mathcal J\xrightarrow{Y}\mathcal C$ so that $X\overset\gamma\Rightarrow Z\circ c$ whiskers as $X\overset\beta\Rightarrow Y\circ b\overset{\alpha . b}\Rightarrow (Z\circ a)\circ b=Z\circ c$.
Choosing $Y=Z\circ a$ and $Y\overset{\mathrm{id}}\Rightarrow Z\circ a$, the unique $X\overset\beta\Rightarrow Y\circ b$ is given by $\beta=\gamma$. In other words, the fibration has the usual splitting of the domain fibration and the $2$-categorical structure plays no role.
