I read in mirror symmetry and algebraic geometry by Cox and Katz that we have stable map $f : C \to Y$ which C is nodal curve with $n$ marked point then we have $0 \to Ext^0_C([f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)],\mathcal{O}_C) \to Ext^0_C(\Omega_C(p_1+p_2+...+p_n),\mathcal{O}_C) \to Ext^0_C(f^*\Omega_Y,\mathcal{O}_C) \to Ext^{1}_C([f^*\Omega_Y \to \Omega_C],\mathcal{O}_C(p_1+p_2+...+p_n)) \to ...$

I guess we can get this long exact sequence by take $Ext$ from short exact sequence

$ 0\to T_C(-p_1-p_2-...-p_n) \to f^*T_X \to [f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]\to 0$

My question is what does $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ mean and how can we get this short exact sequence?

(I know that if $f$ is closed immersion then the last term of short exact sequence is normal bundle but in general we change normal bundle by $[f^*\Omega_Y \to \Omega_C(p_1+p_2+...+p_n)]$ but I don't know why) thanks.