Extending isomorphism of family of pointed curve Suppose that $(X,x)$ be pointed nonsingular curve and we have two families over $X$ of stable maps to$P^r$:
$(\pi:C\longrightarrow X,{p_i},f)$
($(\pi^{'}:C\longrightarrow X,{p_i}^{'},f^{'})$
If we set $U$=$X$\ ${x}$ and two families are isomorph over $U$  i want to prove that we can extend to all of $X$.
First we choose basis ${x_0,...,x_r}$ for $O(1)$ such taht $f:C_x \longrightarrow P^r$ and $f:C^{'}_x \longrightarrow P^r$ intersect transversally at nonsingular,unmarked points then it is claimed that since we can prove this fact over local etale cover of $(X,x)$ we can assume that $f^*(t_i)$ and $f^*{^{'}}(t_i)$ split into sections ${q_i,_j}$ and ${q^{'}_i,_j}$ of $\pi$ and ${\pi}^{'}$.
Why this is true?
 A: I am not addressing the actual proof of separatedness of the stack of Kontsevich stable maps.  Just like for separatedness of the stack of Deligne-Mumford stable curves (studied also by Grothendieck, Knudsen, and Mayer), there is something quite nontrivial in that proof, basically the fact that the "relative log canonical ring" can be recovered from any semistable model that contracts to the stable model (same issue arises for moduli of higher-dimensional objects).  Instead I prove the step that the OP asks about: how to prove there exists a faithfully flat, étale base change that "splits" the inverse images of the coordinate hyperplanes into a union of disjoint sections.
For every ordered basis $T= (t_0,\dots,t_r)$ for $H^0(\mathbb{P}^r,\mathcal{O}(1))$, a family of stable maps to $\mathbb{P}^r$, $$(\pi:C\to X, (p_i:X\to C),f:C\to \mathbb{P}^r)$$ is $T$-balanced if the zero schemes $\text{Zero}(f^*t_i)$ in $C$ are 
*
*pairwise disjoint from one another, 
* disjoint from the images of all of the sections $p_i$, and 
* étale over $X$.   Note, this last condition implies that these zero schemes do not contain any irreducible component of any fiber of $\pi$, and it implies that the zero scheme does not contain any node of any fiber (since otherwise the zero scheme would be nonreduced at that point).  There is an open substack of the stack of stable maps parameterizing $T$-balanced stable maps.  Moreover, for every finite collection of stable maps to $\mathbb{P}^r$, inside the affine variety $\text{Isom}_k(k^{\oplus (r+1)},H^0(\mathbb{P}^r_k,\mathcal{O}(1)))$ of bases $T$, there is a dense Zariski open subset parameterizing those bases $T$ such that each of the finitely many stable maps is $T$-balanced.  Applying this to the two families arising in the proof of separatedness, it suffices to prove separatedness of the open substack of $T$-balanced stable maps.
For every $T$-balanced family of stable maps over a connected base scheme $X$,
define $X_0\subset C$ to be the union of the zero schemes of every $f^*t_i$.  By definition, the restriction of $\pi$ to $X_1$ is étale.  Since $X_0$ is a closed subscheme of $C$, and since $C$ is proper over $X$, also $X_0$ is proper over $X$.  Thus, $X_0$ is finite and étale over $X$.
Lemma.  For every finite and étale morphism of connected schemes, $h:X_0\to X$, there exists a finite, étale, and faithfully flat morphism of schemes $v:\widetilde{X}\to X$ such that the fiber product $\widetilde{X}_0=\widetilde{X}\times_X X_0$ equals the disjoint union of finitely many images of sections of the first projection $\widetilde{X}_0\to \widetilde{X}$.
Proof.  The pushforward of the structure sheaf is a locally free $\mathcal{O}_X$-module.  Since $X$ is connected, this has constant rank, say $r$.  The proof is by induction on $r$, i.e., the length of every fiber of $h$.  If the length $r$ equals $0$, then $X_0$ is empty and there is nothing to prove: $X_0$ is the union of an empty collection of sections.  If the length $r$ equals $1$, then $h$ is an isomorphism.  Then the inverse of $h$ gives a section whose image equals $X_0$.  Thus, by way of induction, assume that $r>1$, and assume the result is proved for $r-1$.  Since $r>0$, the morphism $h$ is also faithfully flat (every locally free sheaf of rank $r>0$ is faithfully flat).
The property of being finite and étale is preserved by base change, so also $X_1 = X_0\times_X X_0$ is finite and étale over $X_0$ of length $r$ with respect to the second projection, $$d_1:X_1\to X_0.$$  Now the diagonal morphism defines a section, $$s_0:X_0\to X_1.$$  Thus, the finite, étale $X_0$-scheme $X_1$ is the disjoint union of the image $s_0(X_0)$ and an open and closed subscheme $X_1'$ that is also finite and étale over $X_0$.  By construction, the length of $X_1'$ as an $X_0$-scheme equals $r-1$.  
By the induction hypothesis, there exists a finite, étale, faithfully flat morphism $\widetilde{X}\to X_0$ such that $\widetilde{X}\times_{X_0} X_1'$ is a disjoint union of images of sections.  Together with the image of the base change of $s_0$, the $\widetilde{X}$-scheme $\widetilde{X}\times_{X_0} X_1$ is a disjoint union of images of sections.  A composition of finite, étale, faithfully flat morphisms is finite, étale, and faithfully flat.  Thus, the composition $\widetilde{X}\to X_0\to X$ is finite, étale, and faithfully flat.  By construction, $$\widetilde{X}\times_X X_0 = \widetilde{X}\times_{X_0} (X_0\times_X X_0) = \widetilde{X}\times_{X_0} X_1.$$  Thus, the $\widetilde{X}$-scheme $\widetilde{X}\times_X X_0$ is a disjoint union of images of sections of the first projection. QED
Now apply the lemma in the case of the $T$-balanced family of stable maps to the union $X_0$ of the zero schemes of all $f^*t_i$.
