We deduce this from the Jacobi-Zariski exact sequence as follows:
Some notation first: Let $H_i(A,B,W)$ where B is an A-algebra and W is a B-module denote $\pi_i(L_{B/A}\otimes_B W)$.
Then given an $A$-algebra $B$, a $B$-algebra $C$, and a $C$-module $W$, we have a long-exact sequence, called the Jacobi-Zariski sequence:
$$\dots \to H_n(A,B,W)\to H_n(A,C,W) \to H_n(B,C,W)\to H_{n-1}(A,B,W)\to \dots \to H_0(B,C,W)\to 0$$.
Now, let $C$ be an $A$-algebra, let $B$ be a polynomial ring over $A$ with an ideal $I$ such that $B/I\cong C$.
Then we have the following long-exact sequence:
$$\dots \to H_n(A,B,C)\to H_n(A,C,C) \to H_n(B,C,C)\to H_{n-1}(A,B,C)\to \dots \to H_0(B,C,C)\to 0$$.
Since $C$ is a quotient of $B$, it is formally unramified over $B$, so $H_0(B,C,C)\cong \Omega_{C/B}=0$. We also see that $I/I^2$ is precisely $H_1(B,C,C)$ by Proposition 1 of Chapter VI.a of the book Homologie des algèbres commuatatives by Michel André. This gives us a long-exact sequence
$$\dots \to H_n(A,B,C)\to H_n(A,C,C) \to H_n(B,C,C)\to H_{n-1}(A,B,C)\to \dots $$
$$\to H_1(A,B,C) \to H_1(A,C,C)\to I/I^2 \to H_0(A,B,C)\to H_0(A,C,C)\to 0$$.
We see that the conormal exact sequence is the truncation at $I/I^2$, and that this sequence splits if $H_0(A,C,C)$ is projective and $H_1(A,C,C)$ is $0$. Conversely, if the sequence is split-exact, then $H_0(A,C,C)$ is a direct summand of $H_0(A,B,C)$, which is projective (it might be free, but I don't remember), and the kernel of $I/I^2\to H_0(A,B,C)$ is $0$, which combined with the fact that $H_1(A,B,C)=0$ implies that $H_1(A,C,C)=0$.
Best Answer
You are right. This is a result due to Van de Ven.
[A. Van de Ven, A property of algebraic varieties in complex projective spaces. In: Colloque Géom. Diff. Globale (Bruxelles, 1958), 151–152, Centre Belge Rech. Math., Louvain 1959. MR0116361 (22 #7149) Zbl 0092.14004]
Even more is true. Recently Ionescu and Repetto proved the following generalization of Van de Ven's Theorem.
Let me sketch a short elementary proof (of Van de Ven's result not its generalization) in the case of hypersurfaces. I will phrase it in the analytic category but once it is translated to the algebraic category, working with infinitesimal neighborhoods, I believe that what will emerge is one of the proofs in the literature.
If the normal sequence splits then we can define a foliation $\mathcal L$ by (germs of) lines everywhere transverse to $X$ at a neighborhood $U$ of $X$. Since the complement of $X$ is Stein we can extend $\mathcal L$ to the whole $\mathbb P^n$. Therefore $\mathcal L$ is defined by a global section of $T \mathbb P^n(d-1)$ for some $d \ge 0$. With the help of Euler's sequence, this section can be presented as a homogeneous vector field $v$ on $\mathbb C^{n+1}$ with coefficients of degree $d$. To compute the tangencies between $\mathcal L$ and $X$ we have just to contract the differential $dF$ of a defining equation $F$ of $X$ with $v$. If $F$ is not linear then the divisor on $X$ defined by the tangencies between $\mathcal L$ and $X$ (defined by $F=dF(v)=0$) will be non-empty contradicting the transversality between $X$ and $\mathcal L$.