2. As shown in Figure 1, if quadrilateral $A B C D$ is folded along $E F$, such that points $A$ and $B$ fall inside quadrilateral $E F C D$, there is a constant quantitative relationship between $\angle 1+\angle 2$ and $\angle A$, $\angle B$. Then, we have $(\quad$.
(A) $\angle 1+\angle 2=(\angle A+\angle B)-90^{\circ}$
(B) $\angle 1+\angle 2=2(\angle A+\angle B)-90^{\circ}$
(C) $\angle 1+\angle 2=2(\angle A+\angle B)-180^{\circ}$
(D) $\angle 1+\angle 2=2(\angle A+\angle B)-360^{\circ}$
2. D.

As shown in Figure 4, let points $A$ and $B$ fall on points $A^{\prime}$ and $B^{\prime}$, respectively. Extend $E A$, $F B$, $E A^{\prime}$, and $F B^{\prime}$ to intersect at points $P$ and $P^{\prime}$, and connect $P P^{\prime}$. By symmetry, we have
$$
\begin{array}{l}
\angle E P P^{\prime}=\angle E P^{\prime} P, \\
\angle F P P^{\prime}=\angle F P^{\prime} P .
\end{array}
$$

Notice that
$$
\begin{array}{l}
\angle 1=\angle E P P^{\prime}+\angle E P^{\prime} P=2 \angle E P P^{\prime}, \\
\angle 2=\angle F P P^{\prime}+\angle F P^{\prime} P=2 \angle F P P^{\prime},
\end{array}
$$
$$
\begin{array}{l}
\text { then } \angle 1+\angle 2=2\left(\angle E P P^{\prime}+\angle F P P^{\prime}\right)=2 \angle C P D . \\
\text { Also, }\left(180^{\circ}-\angle A\right)+\left(180^{\circ}-\angle B\right)+\angle C P D=180^{\circ},
\end{array}
$$

which means $\angle A+\angle B=180^{\circ}+\angle C P D$.
Therefore, $\angle 1+\angle 2=2(\angle A+\angle B)-360^{\circ}$.