The answer depends very much on the geometric generic fiber of $C$. Assume first that the geometric generic fiber of $C$ has genus $\geq 1$. Denote by $$\mathcal{C} \subset \mathbb{P}^2_{ \overline{ \mathbb{Q} } } \times_{ \overline{ \mathbb{Q} } } \mathbb{P}^2_{ \overline{ \mathbb{Q} } }$$ the Zariski closure of $C$, i.e., the generic fiber of projection onto the first factor, $$\pi_1:\mathcal{C} \to \mathbb{P}^2_{ \overline{ \mathbb{Q} } },$$ equals $C$. The morphism $\pi_1$ is flat away from a finite set $J$ of closed points of $\mathbb{P}^2_{ \overline{ \mathbb{Q} } }$. Moreover, $\pi$ is smooth away from a proper closed subscheme $Z\subset \mathbb{P}^2_{ \overline{ \mathbb{Q} } }$ By Weil's extension theorem together with the Abel map from $C$ into (an appropriate torsor for) its Jacobian, every $I_P$ is contained in $Z$, cf. Theorem 1, p. 109 of "Néron Models" by Bosch, Lütkebohmert and Raynaud. Moreover, there is a Néron model of this (torsor for the) Jacobian at codimension $1$ points contained in $Z$, the Abel map extends to the smooth locus of these fibers, and it may extend to singular points of the fiber as well (depending on the type of the fiber). So you may be able to further limit $I_P$.

However, if $C$ is a genus $0$ curve, all bets are off.

**Edit**. Also, if $g>1$ and if the family is non-isotrivial, then there are only finitely many points $P$ in the first place (by Mordell for function fields). Also, if $g>1$, even if the family is isotrivial, nonetheless $f$ is constant. So again the union of $I_{f^n(p)}$ is finite. So the "real" cases are $g=1$, i.e., plane cubics, and $g=0$, i.e., lines and conics.

(There was more here, but it was wrong. The Néron extension property for the Néron model really says nothing about where in $Z$ the indeterminacy can be, except that it cannot contain any points over which the Abel map extends.)