Let $p_1,p_2:X\times X\to X$ be the two projections to $X$. Then $\Omega_{X\times X}\simeq p_1^*\Omega_X\oplus p_2^*\Omega_X$ and the residue map $\Omega_{X\times X}\otimes \mathscr O_{\Delta}\to \Omega_\Delta$ induces an isomorphism $p_1^*\Omega_X\otimes \mathscr O_{\Delta}\simeq \Omega_\Delta$, since $p_1|_\Delta$ is an isomorphism.Then it is easy to see that the kernel of this map is isomorphic to $p_2^*\Omega_X\otimes \mathscr O_{\Delta}\simeq \Omega_\Delta$.
So, does this mean that it should be geometrically intuitive that indeed the co-normal bundle of the diagonal is isomorphic to the cotangent sheaf? I say yes: If you repeat this argument in plain words it says that the (co)tangent sheaf of $X\times X$ restricted to the diagonal is naturally (cf. what does "natural" mean?) the direct sum of two copies of the (co)tangent sheaf of $\Delta\simeq X$. Furthermore, when we decompose this the restriction of (co)tangent vectors of $X\times X$ to (co)tangents co)tangent and (co)normals, co)normal vectors of $\Delta$, then since the first is what it is(co)tangent of $\Delta$ can be identified with one of the two components of the (co)tangent of $X\times X$, the second should also be that(co)normal gets identified with the other.
I would add one more important issue that can get lost both in an intuitive and a local argument. It is important to keep track of what maps where naturally. Verifying an isomorphism of sheaves locally only works if one already has a morphism globally. In particular, in Michael's answer, when he says "...the normal is the quotient...", he is using the fact that there is such a map globally that restricts to this one. Otherwise one would only prove that there is a local isomorphism, which between locally free sheaves only means that they have the same rank. In other words, to do this correctly you need to argue with the (global) short exact sequence. Then again, if you truly understand that sequence, it is equivalent to the intuitive argument above about (co)tangents.
