Which rank 1 bundle over $\mathbb{P}^1$ is this exceptional divisor? Take two intersecting lines $L_1,L_2$ in $\mathbb{P}^3$. Blow-up $L_1$, and then blow up the strict transform $L_2'$ of $L_2$ and call $E_2$ the exceptional divisor of this second blow-up. $E_2$ is a $\mathbb{P}^1$-bundle over $L_2'\cong\mathbb{P}^1$ and so is some Hirzebruch surface $F_n$.
By doing local computations, I think that $E_2$ is $F_1$. Is there a good way (I mean, without coordinates, or at least without having to use affine charts) to figure out which $F_n$ it is?
 A: Let $\pi \colon X \to \mathbb{P}^3$ be the first blow-up. 
The normal bundle of $L_2$ in $\mathbb{P^3}$ is $$N_{L_2 / P^3}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}(1);$$ in fact $L_2$ moves into a family of dimension $4=h^0(N_{L_2/ P^3})$, namely the projective Grassmannian $\mathbb{G}(1,3)$.
The strict transform $L_2'$ of $L_2$, instead, moves into a family of smaller dimension, namely the strict transform of the family $\mathfrak{X}$ of lines intersecting $L_1$. Since $\mathfrak{X}$ is a divisor in $\mathbb{G}(1,3)$, we have $h^0(N_{L_2'/ X})=3$. In fact, one proves that 
$$N_{L_2' / X}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1},$$
see the edit at the endo of the post.
It follows that the exceptional divisor of the blow-up of $X$ along $L_2'$ is the projective bundle $$\mathbb{P}(\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}),$$
which is in turn isomorphic to the Hirzebruch surface $\mathbb{F}_1$.
EDIT. Let me prove in full details that  $N_{L_2' / X}=\mathcal{O}_{P^1}(1) \oplus \mathcal{O}_{P^1}.$
Let us consider the family $\mathcal{S}$, given by the strict transforms in $X$ of the planes containing $L_2$, and let $S$ be a general element of $\mathcal{S}$. Then $S$ is isomorphic to $\mathbb{F}_1$, since it is just a plane blown-up at the point $L_1 \cap L_2$. Moreover $L_2' \subset S$ is a fibre of the ruling. Finally, since two elements of $\mathcal{S}$ intersect in a reducible curve made by a fibre and the $(-1)$-curve, it follows $$N_{S/X}=\mathcal{O}_S(S) \cong \mathcal{O}_{\mathbb{F}_1}(C_0+f),$$
where $C_0$ is the $(-1)$-curve and $f$ is a fibre of the ruling (of course, these $\mathbb{F}_1$ have nothing to do with the exceptional divisor of the second blow-up...)
Now, by using the normal bundle sequence associated with $L_2' \subset S \subset X$ we obtain
$$0 \to N_{L_2' / S} \to N_{L_2' / X} \to N_{S / X} \otimes \mathcal{O}_{L_2'} \to 0,$$
that is
$$ 0 \to \mathcal{O}_{P^1} \to N_{L_2' / X} \to \mathcal{O}_{P^1}(1) \to 0$$
(the last line bundle on the right comes from the fact that $(C_0+f)f=1$).
Since $\operatorname{Ext}^1(\mathcal{O}_{P^1}(1), \mathcal{O}_{P^1})=H^1(\mathcal{O}_{P^1}(-1))=0,$ the last exact sequence splits and the claim follows.
A: Here is an alternative approach without referring to deformations.
Let $P=\mathbb P^3$, $B$ denote the first blow up, $E$ the exceptional divisor of the blow up $\pi:B\to P$, and $L$ the proper transform of $L_2$. Also let $k$ denote the base field.
Then $E\simeq \mathbb{P^1\times P^1}$ and $\Omega_{B/P}$ is supported on $E$ where it is isomorphic to a line bundle (actually $\mathscr O_{E/L_1}(-2)$, but this does not matter). So, we have that $$\Omega_{B/P}\otimes \mathscr O_L\simeq k.$$
On $B$ there is a short exact sequence
$$ 0\to \pi^*\Omega_P \to \Omega_B \to \Omega_{B/P} \to 0$$
which remains exact after restricting to $L$:
$$ 0\to \pi^*\Omega_P\otimes \mathscr O_L \to \Omega_B\otimes \mathscr O_L \to \Omega_{B/P}\otimes \mathscr O_L \to 0$$
Then there is the conormal sequence of $L$ in $B$:
$$
0\to \mathscr N_{L/B}^{\ \ \vee} \to \Omega_B\otimes \mathscr O_L \to \Omega_L \to 0,
$$
and there is a similar short exact sequence on $L_2$, which we pull back to $L$:
$$
0\to \pi^*\mathscr N_{L_2/P}^{\ \ \vee} \to \pi^*\Omega_P\otimes \mathscr O_{L} \to \pi^*\Omega_{L_2} \to 0.
$$
Notice that the second short exact sequence maps to the first and that 
$\pi$ is an isomorphism between $L$ and $L_2$, so using the previous short exact sequence and the Snake Lemma, we obtain that there is a short exact sequence:
$$
0\to \pi^*\mathscr N_{L_2/P}^{\ \ \vee}\to \mathscr N_{L/B}^{\ \ \vee}  \to k\to 0.
$$
Finally, $\pi^*\mathscr N_{L_2/P}^{\ \ \vee}\simeq \mathscr O_L(-1)\oplus\mathscr O_L(-1)$, so it follows that $\mathscr N_{L/B}^{\ \ \vee}\simeq \mathscr O_L(a)\oplus\mathscr O_L(b)$ such that $a,b\geq -1$ and $a+b=-1$. This can only happen if
$$\mathscr N_{L/B}\simeq \mathscr O_L\oplus\mathscr O_L(1),$$
and then blowing up $L$ the exceptional divisor is the Hirzebruch surface $\mathbb F_1$.
A: Here is another approach, not even referring to exact sequences. Let's do this torically - using Fulton's fan language. The fan of ${\mathbb P}^3$ is given by the cones spanned by the vectors $e_1,e_2,e_3,-e_1-e_2-e_3$ in the vector space spanned by them. The first blowup corresponds to subdividing the fan by inserting the vector $e_1+e_2$, whereas the second blowup corresponds to subdividing the fan further by inserting the vector $e_1+e_3$ - please draw a picture! Staring at the picture will tell you that this last vector will give an edge of the four cones $(e_1, e_1+e_2, e_1+e_3)$, $(e_1+e_2, e_3, e_1+e_3)$,  $(e_3, -e_1-e_2-e_3, e_1+e_3)$, $(-e_1-e_2-e_3, e_1, e_1+e_3)$. The geometry of the divisor corresponding to the vector $v=e_1+e_3$, the exceptional divisor $E_2$ you are looking for, is given by projecting these four cones onto the quotient of the vector space by the span of $v$; you can arrange this simply by setting $e_1+e_3=0$ in the above cones, getting the cones $(e_1, e_1+e_2)$, $(e_1+e_2, -e_1)$,  $(-e_1, -e_2)$ and $(-e_2, e_1)$ in two dimensions. This is the standard fan of ${\mathbb F}_1$. 
