I've a confusion regarding Corollary 2.4. It says that the graphs $\Gamma_f$, $\Gamma_g$ gives us sections which agree at $y'$. I don't understand what section it is talking or it is a section of which map? Also, why does that imply $\Gamma_f$, $\Gamma_g$ are isomorphic?
It means that $\Gamma_f,\Gamma_g: Y'\to Y'\times_X Y$ are sections of $Y'\times_X Y\to Y'$, and hence thanks to Lemma 2.3 (you need $Y'$ connected, which is not explicit in the hypothesis) they give you isomorphisms between $Y'$ and connected components $U_f, U_g$ of $Y'\times_X Y$. But, since $f(y')=g(y')$ and the two maps $k(y)\to k(y')$ agree, this tells you that the two maps ${\rm Spec}~k(y')\to Y$ agree and hence, by the usual property of the fiber product, also the two maps ${\rm Spec}~ k(y')\to Y'\times_X Y$ agree and hence $\Gamma_f(y')=\Gamma_f(y)$, which implies $U_f=U_g=U$.
Now, $f$ is equal to the composition $Y'\to U_f\to Y'\times_X Y\to Y$, and the same holds for $g$. Hence, to show that $f=g$, you need to show that the two isomorphisms $Y'\to U_f=U$ and $Y'\to U_g=U$ are the same. But this is obvious, since they are both inverses of the projection $U\subseteq Y'\times_X Y\to Y'$.