Computing the euler number of an oriented disc bundle over a sphere Suppose $E\to S^2$ is an (smooth) oriented disk bundle. Let $D_1$ and $D_2$ denote the upper and lower hemispheres of $S^2$, respectively. Then $E|_{D_1}$ and $E|_{D_2}$ are trivial, so they are both diffeomorphic to $D_i\times D^2$. So $E$ may be seen to be obtained from $D_1\times D^2$ and $D_2\times D^2$ by attaching their boundaries, for some diffeomorphism $f:\partial D_1 \times D^2 \to \partial D_2\times D^2$ which is identity on the first factor, and an element of $SO(2)$ on the second factor. (Thus $f(x,y)=(x,g(x)y)$ for some $g:\partial D_1\to SO(2)$. Thus $g$ gives an element $\pi_1(SO(2))\cong \Bbb Z$, say $n$. We may assume $g$ is the $n$-fold product of the standard generator of $\pi_1(SO(2))$.
Note that $E$ is homotopy equivalent to the zero section $\cong S^2$. I want to show that the euler number of $E$ is $n$, by showing that the self intersection number of the zero section is $n$. To this end, it suffices to show that there is a section $h:D_1\cup D_2\to E=D_1\times D^2 \cup_f D_2\times D^2$ which intersects the zero section $|n|$ times positively if $n>0$ (negatively if $n<0$). First, fix an arbitrary point $p(\neq 0)\in D^2$ and define $h(x)=(x,p)\in D_1\times D^2$ for $x\in D_1$. Then we must have $h(x)=(x,g(x)p)\in D_2\times D^2$ for $x\in \partial D_2$. Now extend $h$ over $D_2$ by defining $h(rx)=(rx,r^2g(x)p+(1-r^2)q)$ where $q=(1,0)\in D^2$ and $0\leq r\leq 1$. This gives a section of $E\to S^2$ that intersects the zero section exactly $n$ times, but how can we know whether the intersection is positive or negative?
Or is there another method for computing the euler number of $E$?
P.S. Does every oriented $D^2$-bundle over $S^2$ is the unit disk bundle of some complex line bundle over $S^2$ with a hermitian metric?
 A: The Euler class depends homomorphically on $n$ so it suffices to calculate it for $n=1$. In this case, the disk bundle we get is fiberwise diffeomorphic to the disk bundle of the tautological bundle over $\mathbb{C}P(1)$. The Thom space of this bundle is $\mathbb{C}P(2)$. The self intersection number can be calculated via the cup square. The zero section corresponds to the inclusion of $\mathbb{C}P^1$ into $\mathbb{C}P^2$ and so the cup square of the dual of this class is the fundamental class by the cohomology structure. This means the Euler number is 1.
Hence, for general $n$ the Euler number will be $n$.
To address your other question: I highly recommend figuring this issue with orientations out yourself. It will be very instructive.
And yes this follows from the fact that $\pi_2(BO(2))$ is generated by the tautological bundle over $\mathbb{C}P^1$. If one wanted to do this calculation via complex means, we could use the fact that the Chern number of the bundle is 1 and that this is equal to its Euler number.
