Equivalence between wavefunction and Dirac ket notation I am a bit puzzled about how different notations can be used interchangeably in quantum mechanics.
For example, the time-independent Schrodinger equation (without potential) can be written in two slightly different forms:
$$
-\frac{\hbar^2}{2m}\Delta\ |\Psi\rangle = E\ |\Psi\rangle
\\
or
\\
-\frac{\hbar^2}{2m}\Delta\ \Psi(\textbf{r}) = E\ \Psi(\textbf{r})
$$
These two notations are considered equivalent.
But they are not stricly the same, since $$|\Psi\rangle \neq \Psi(\textbf{r})$$
Indeed, the actual relation is: $$|\Psi\rangle = \int\limits_{n} d^n\textbf{r}\ \Psi(\textbf{r}) | \mathbf{r} \rangle$$
If I see $|\Psi\rangle$ as $\sum\limits_i \Psi_i |u_i\rangle$ (which is kind of the discrete form of the last equation, right?), I can understand intuitively why $|\Psi\rangle$ and $\Psi(\textbf{r})$ can be used interchangeably. But I'd like to know if there is a more rigourous way to explain it. 
 A: Let me show every step necessary to convert between the two notations. Your first equation is properly written as 
$$\hat{H} \, | \psi \rangle = E\,  | \psi \rangle$$
where $\hat{H}$ is an operator, $|\psi \rangle$ is a state vector, and $E$ is a number. Now, both $\hat{H}$ and $|\psi \rangle$ may be expressed in terms of components in a basis. For example, in the position basis, the components of $|\psi \rangle$ form a function, called the wavefunction,
$$\langle x | \psi \rangle = \psi(x).$$
The Hamiltonian is an operator, so in components, it becomes a matrix, with components
$$H_{xy} = \langle x | \hat{H} | y \rangle.$$
Now let's apply this to the first equation. We hit everything on the left with $\langle x |$, and also insert a copy of the identity,
$$1 = \int dy |y \rangle \langle y|$$
on the left-hand side. This gives
$$\int dy\, \langle x | \hat{H} | y \rangle \langle y | \psi \rangle = E \langle x | \psi \rangle.$$
Expanding in components, we have
$$\int dy \, H_{xy} \psi(y) = E \, \psi(x).$$
This is the Schrodinger equation in components. In your specific case, the components are
$$H_{xy} = -\frac{\hbar^2}{2m} \delta''(x-y).$$
That is, the components are the second derivative of a Dirac delta function. Plugging this in and integrating by parts twice, we have
$$- \frac{\hbar^2}{2m} \int dy \, \delta(x-y) \psi''(y) = E\,  \psi(x)$$
Performing the integral, we have
$$-\frac{\hbar^2}{2m} \psi''(x) = E \psi(x).$$
Finally, we can write this in the "abstract" notation 
$$- \frac{\hbar^2}{2m} \nabla^2 \psi = E \psi$$
which is your second equation. The difference is that the first equation is truly an abstract operator equation, independent of basis. In the second equation, we have separated out $H$ as an operator, but it acts on coefficients, not on the state vectors themselves. As such, this equation only is useful when working in a particular basis, the position basis.

Emilio Pisanty has given a nice (and much shorter) answer. I think it's good to see the full computation a few times, but after that, you really don't want to descend all the way to components like I just did, since component expressions for operators tend to be very ugly.
A: The formulas you write are not quite there. The connections between the wavefunction and the state vector are
$$
|\Psi\rangle = \int \mathrm d\mathbf r \:\Psi(\mathbf r)|\mathbf r\rangle
$$
and
$$
\Psi(\mathbf r) = \langle \mathbf r | \Psi \rangle.
$$
Mixing up the derivative notation with Dirac notation is also a pretty bad idea and rather liable to get you confused pretty soon. Instead, when using Dirac notation, it's more common to denote the kinetic term in terms of the momentum operator, and the way to relate the two via Dirac notation is as
$$
\langle \mathbf r|  \hat{\mathbf p}^2 = -\hbar^2 \nabla^2 \langle \mathbf r|.
$$
Thus, if you take the Dirac-notation Schrödinger equation
$$
\frac{\hat{\mathbf p}^2}{2m}|\Psi\rangle = E|\Psi\rangle
$$
and you multiply it on the left by $\langle\mathbf r|$, you get the Schrödinger equation for the wavefunction,
$$
-\frac{\hbar^2}{2m}\nabla^2\Psi(\mathbf r) = E\Psi(\mathbf r).
$$
