Let $\mathscr A$ be a category and $\mathbf I$ a small category. Let $D:\mathbf I\to\mathscr A$ be a diagram in $\mathscr A$, and write $D^{op}$ for the corresponding functor $\mathbf I^{op}\to\mathscr A^{op}$. A cocone on $D$ is a cone on $D^{op}$, and a colimit of $D$ is a limit of $D^{op}$.
Explicitly, a cocone $D$ is an object $A\in\mathscr A$ together with a family $(f_I:D(I)\to A)_{I\in\mathbf I}$ of maps in $\mathscr A$ such that for all maps $u:I\to J$ in $\mathbf I$, $f_J\circ Du=f_I$.
I don't understand how exactly it follows from the definition of a cone on $D^{op}$ that it is a family $(f_I:D(I)\to A)_{I\in\mathbf I}$ (as opposed to $(f_I:A\to D(I))_{I\in\mathbf I}$. To begin with, the notion of a cone on a functor $F$ has previously been defined (Definition 5.1.19) only for functors $F$ from $\mathbf I$ to $\mathscr A$. How did the author obtain the quoted part starting with "Explicitly,..." then, given that no definition of a cone on a functor $\mathbf I^{op}\to\mathscr A$ has been given in the first place?
Update: Okay, probably we don't need a separate definition of a cone on a functor with domain $\mathbf I^{op}$, but I still don't have a lucid understanding of how to obtain the explicit form using the definitions. The significance of using $D^{op}$ shows up when we take images of $f^{op}:J\to I$ under $D^{op}$, but when talking about the family $(f_I:D(I)\to A)_{I\in\mathbf I}$, we are not talking about arrows $J\to I$ (later, when describing the property, we do, but a priori not).
A cocone on $D$ is a cone on $D^{op} : \mathbb{I}^{op} \to \mathscr{A}^{op}$, that is, it is an object $A \in \mathscr{A}^{op}$ together with a family of maps $(f_{I}: A \to D^{op} (I))_{I \in \mathbb{I}^{op}}$ in $\mathscr{A}^{op}$.
So the maps $f_I$ are in the opposite category $\mathscr{A}^{op}$. The opposite cateogry $\mathscr{A}^{op}$ has the same objects as $\mathscr{A}$ but the direction of the morphisms is reversed. So $f_{I}: A \to D^{op} (I)$ in $\mathscr{A}^{op}$ is the same as saying $f_{I}: D^{op}(I) \to A$ in $\mathscr{A}$. Also, $D^{op} (I) =D(I)$.