Why is this an equivalent condition for stability of curves mapped to projective space? In Fulton-Pandharipande's Notes on Stable Maps and Quantum Cohomology he claims on page 11 that if $X = \mathbb{P}^r$, the stability of a flat family of curves $(\pi:C\to S, \{p_i\}, \mu)$ where
$$
\mu: C \to \mathbb{P}^r_S
$$
is equivalent to having $\omega_{C/S}(p_1+\cdots+p_n)\otimes \mu^*(\mathcal{O}_{\mathbb{P}_S^r}(3))$ is $\pi$-ample. Why is this true?
 A: We want to show that stability of an individual map $(C,\left\{p_i\right\},\mu:C \to \mathbb P_k^r)$ in the family is equivalent to ampleness of 
$$\omega_C(p_1 + \cdots + p_n)\otimes \mu^*(\mathcal O_{\mathbb P_k^r}(3))$$ on the curve $C$. I'll do one direction. If you know a bit about the basic theory of stable  curves, you know that a pointed curve is stable iff
$$
\omega_C(p_1 + \cdots + p_n).
$$
is ample (which is the same as "of positive degree" on a curve, and can be checked component-wise in the reducible case). Looking at the definition of stable map, we see that the domain curve need not be stable, but any components on which the Deligne-Mumford stability condition fails (i.e. isolated unmarked genus $1$ curves, or genus $0$ curves with fewer than $3$ special points) must not map to points in $\mathbb P^r$. But then the image of such a component will intersect a general cubic hypersurface in $\geq 3$ points, so $\mu^*\mathcal O(3)$ will be ample on this component.
So we have two line bundles, one of which is ample on each component. On those which are not stable as curves, $\deg\mu^*\mathcal O(3) \geq 3$, and since twisting by $\omega_C(p_1 + \cdots + p_n)$ can decrease the degree by at most $2$ on a given component (in the case of unpointed $\mathbb P^1$), we are still good.
Now on any component which was DM stable to begin with, we need only confirm that twisting by $\mu^* \mathcal O(3)$ does not decrease the degree, but this is clear for any number of reasons (e.g. nefness is preserved under pullback via morphisms, and nef on a curve is the same as "of nonnegative degree").
The other direction of the equivalence should also follow from the stable curve version.
