Let $x\in X=\mathbb{P}^3$. Consider the blow-up $\widetilde{X}\to X$ of $x$ in $X$. Let $l\ni x$ be a line in $X$ and $\widetilde{l}$ is its strict transform in $\widetilde{X}$. How to prove that the normal bundle $N_{\widetilde{l}/\widetilde{X}}$ is $\mathcal{O}\oplus \mathcal{O}$?
Upd. Let us blow two points on $l$. How to prove that the normal bundle of strict transform would be $\mathcal{O}(-1)\oplus \mathcal{O}(-1)$?
(Though it's arguably poor taste to make such extensive edits, particularly after an answer has been up-voted, in this case it seemed justified.)
First, here's an intuitive geometric argument: The space of lines through the point $x$ is a copy of $\mathbf{P}^2$. If $l$ and $l'$ are distinct lines through $x$, their strict transforms in the blow up are disjoint (more or less obvious from the definition of a blow up). So, there is a (complex) $2$-dimensional family of perturbations of $\tilde{l}$ in $\widetilde{X}$, and an arbitrary perturbation moves $\tilde{l}$ completely off itself; the normal bundle of $\tilde{l}$ in $\widetilde{X}$ must therefore be trivial.
To flesh out details of my late-night sketch:
Pick a hyperplane $H$ not containing $x$. Projection away from $x$ defines a holomorphic map $\pi:\mathbf{P}^3\setminus\{x\} \to H$ whose fibres are lines through $x$; this identifies the total space of $\mathcal{O}_{H}(1)$, the normal bundle of $H$ in $\mathbf{P}^3$, as the complement of $x$ in $\mathbf{P}^3$.
Consider the "completion" of $\mathcal{O}_H(1)$, i.e., the ruled $3$-fold $M = \mathbf{P}\bigl(\mathcal{O}_H(1) \oplus \mathcal{O}\bigr) \to H$. As a $\mathbf{P}^{1}$ bundle, $M$ is obtained from the principal $\mathbf{C}^\times$ bundle $\mathcal{O}_H(1)^\times$ by the standard action $(\zeta, z) \in \mathbf{C}^\times \times \mathbf{P}^1 \mapsto \zeta z \in \mathbf{P}^1$. From this description, we read off that the normal bundle of the zero section is $\mathcal{O}_H(1)$ (as expected), and the normal bundle of the infinity section $E$ is $\mathcal{O}_E(-1)$, since $\mathbf{P}\bigl(\mathcal{O}_H(1) \oplus \mathcal{O}\bigr) = \mathbf{P}\bigl(\mathcal{O} \oplus \mathcal{O}_E(-1)\bigr)$, i.e. $E$ is the zero section in the $\mathbf{P}^1$ bundle induced by the dual action $(\zeta, z) \mapsto \zeta^{-1} z$.
Particularly, the infinity section $E \subset M$ is biholomorphic to $H$, and the normal bundle of $E$ in $M$ is $\mathcal{O}_E(-1)$. Consequently, $M$ can be blown down by contracting $E$ to a point, and the blow down is $\mathbf{P}^3$ since the complement of $E$ in $M$ is the total space of $\mathcal{O}_H(1)$, i.e., $\mathbf{P}^3\setminus\{x\}$; that is, $M$ is biholomorphic to $\widetilde{X}$.
The strict transform $\tilde{l}$ is a fibre of $\pi:M \to H$ (projection away from $x$ maps $l$ to a point). The normal bundle of $\tilde{l}$ in $\widetilde{X} = \mathbf{P}\bigl(\mathcal{O}_H(1) \oplus \mathcal{O}\bigr)$ is therefore $\mathcal{O} \oplus \mathcal{O}$ since a fibre bundle is locally trivial and $H$ is a manifold.
To address your new question briefly, the normal bundle of $l$ in $\mathbf{P}^3$ is $\mathcal{O}_{\tilde{l}}(1) \oplus \mathcal{O}_{\tilde{l}}(1)$. Blowing up a point of $l$ tensors the normal bundle with $\mathcal{O}_l(-1)$. Blowing up a second point of $l$ (not on the exceptional divisor) tensors the normal bundle by $\mathcal{O}_l(-1)$ again. Alternatively, you should be able to do a local coordinates calculation by blowing up the "origin" in $\mathbf{P}^1 \times \mathbf{C}^2$ and looking at the topological self-intersection of the proper transform of $\mathbf{P}^1 \times \{(0, 0\}$.
