The wavefunction contains all the information about the system, but it is not itself observable. Since our intuition tends to be linked to what we have observed in the past, it's not surprising that the wavefunction seems unintuitive. As you say it can be negative, but it can also be a complex number making it even less intuitive (if that's possible!).
Anything we can observe is obtained from the wavefunction using a function called an operator. In your example of calculating the probability of finding the electron in some infinitesimal volume $dV$, this probability is given by:
$$ P = \psi^*\psi dV $$
where $\psi^*$ is the complex conjugate of $\psi$. The product $\psi^*\psi$ is always real and positive, so the probability of finding an electron is always real and positive. We don't need to worry that the wavefunction can be negative or complex.
The time-independent Schroedinger equation for the hydrogen atom is
$$ H \ \Psi(\vec r_N, \vec r_e) = E \ \Psi(\vec r_N, \vec r_e)$$
where $\vec r_N$ is the position of the nucleus, $\vec r_e$ the position of the electron and
$$H = \frac{p_N^2}{2 m_N} + \frac{p_e^2}{2 m_e} - \frac{e^2}{4 \pi \epsilon_0 \mid \vec r_N - \vec r_e\mid} \, .$$
We usually use center of mass coordinates:
$$\vec r = \vec r_e - \vec r_N\\\vec R= \frac{m_N\ \vec r_N + m_e \vec r_e}{m_N+m_e}$$
so that the Schroedinger equation becomes
$$H \ \Psi(\vec R,\vec r) = E \ \Psi (\vec R,\vec r)$$
where the hamiltonian is now
$$H = \frac{P^2}{2 M} + \frac{p^2}{2 \mu} + \frac{e^2}{4 \pi \epsilon_0 \mid \vec r\mid} $$
where $M=m_N+m_e$ is the total mass, $\vec P = \vec p_N + \vec p_e$ is the total momentum, $\mu \equiv (m_N m_e)/(m_N+m_e)$ is the reduced mass and $\vec p = (m_N \vec p_e - m_e \vec p_N)/(m_N+m_e)$ is the relative momentum.
The total wave function is the product of the nuclear and the electronic wave function ($\Phi$ and $\psi$ respectively):
$$\Psi(\vec R, \vec r) = \Phi(\vec R) \ \psi(\vec r)$$
where
$$\frac{P^2}{2M} \ \Phi (\vec R) = E_{CM} \ \Phi (\vec R)$$
and
$$\left(\frac{p^2}{2 \mu} + \frac{e^2}{4 \pi \epsilon_0 \mid \vec r\mid}\right)\ \psi (\vec r)= E \ \psi (\vec r) \, .$$
The orbitals we always talk about and that you have seen depicted (see following picture for example) are the $\psi(\vec r)$ and are usually expressed in polar coordinates: $\psi(r,\theta,\phi)$.
So what we usually refer to as "orbital" is not the wave function of the whole atom (which would be $\Psi$) nor the electronic wave function, but the wave function of a fictitous particle with mass $\mu$ , position $\vec r$ and momentum $\vec p$.
However, since $m_N \simeq 1800 \ m_e$, you can see that $\vec R \simeq \vec r_N$, $M \simeq m_N$ and $\mu \simeq m_e$, so that $\Phi$ is in a certain way close to the nuclear wave function and $\psi$ to the electronic wave function. Also notice that the Schroedinger equation for $\Phi$ is a free particle equation, which is trivial to solve and will held a plane wave solution.
Best Answer
It's not quite true that "the square of the wave function gives the probability of finding an electron at a particular position." The square of the wave function gives you the probability density of find the the electron as a particular position. IN order to get the probability of finding the electron in a given region of space, you need to integrate the square of the (normalized) wave function over the volume you're interested in.
Another way to think about why you need to integrate over some volume is that the wave function is giving you the probability density at a given point in space. That point is infinitesimally small; in mathematical terms, it has measure zero. Because it is infinitesimally small, there is an infinitesimally small probability of finding the electron there. You need to integrate over a finite volume in order to get a finite probability.