The answers so far seem pretty good, but I'd like to try a slightly different angle.
Before I get to atomic orbitals, what does it mean for an electron to "be" somewhere? Suppose I look at an electron, and see where it is (suppose I have a very sophisticated/sensitive/precise microscope). This sounds straightforward, but what did I do when I 'looked' at the electron? I must have observed some photon that had just interacted with that electron. If I want to get an idea of the motion of the electron (no just its instantaneous momentum, but its position as a function of time), I need to observe it for a period of time. This is a problem, though, because I can only observe the electron every time it interacts with a photon that I can observe. It's actually impossible for me to observe the electron continuously, I can only get snapshots of its position.
So what does the electron do between observations? I don't think anyone can answer that question. All we can say is that at one time the electron was observed at point A, and at a later time it was observed at point B. It got from A to B... somehow. This leads to a different way of thinking about where an electron (or other particle) is.
If I know some of the properties of the electron, I can predict that I'm more likely to observe an electron in some locations than in others. Atomic orbitals are a great example of this. An orbital is described by 4 quantum numbers, which I'll call $n$, $l$, $m$, $s$ (there are several notations; I think this one is reasonably common). $n$ is a description of how much energy the electron has, $l$ describes its total angular momentum, $m$ carries some information about the orientation of its angular momentum and $s$ characterizes its spin (spin is a whole topic on its own, for now let's just say that it's a property that the electron has). If I know these 4 properties of an electron that is bound to an atom, then I can predict where I am most likely to observe the electron. For some combinations of $(n,l,m,s)$ the distribution is simple (e.g. spherically symmetric), but often it can be quite complicated (with lobes or rings where I'm more likely to find the electron). There's always a chance I could observe the electron ANYWHERE, but it's MUCH MORE LIKELY that I'll find it in some particular region. This is usually called the probability distribution for the position of the electron. Illustrations like these are misleading because they draw a hard edge on the probability distribution; what's actually shown is the region where the electron will be found some high percentage of the time.
So the answer to how an electron "jumps" between orbitals is actually the same as how it moves around within a single orbital; it just "does". The difference is that to change orbitals, some property of the electron (one of the ones described by $(n,l,m,s)$) has to change. This is always accompanied by emission or absorption of a photon (even a spin flip involves a (very low energy) photon).
Another way of thinking about this is that the electron doesn't have a precise position but instead occupies all space, and observations of the electron position are just manifestations of the more fundamental "wave function" whose properties dictate, amongst other things, the probability distribution for observations of position.
Yes, atomic orbitals are very significant.
An electron being in a particular orbital corresponds to a specific energy. If an electron transitions between two orbitals, the energy of the photon absorbed or emitted is the difference between the energy levels of the orbitals.
The probablilty density function is also important. For example, an s-orbital electron has a probabilty of being within the nucleus. The gives rise to Fermi Contact Interaction, which has observable effects in NMR and ESR spectroscopy and electron capture.
Best Answer
The hydrogen atom is spherically symmetric, so for any solution of the Schrödinger equation for the hydrogen atom, any rotation of that solution must also be a solution. If you do the math on how to rotate a solution, it turns out that the solutions with a particular energy $E_n$ fall into groups labeled by an integer $l < n$. The integer $l$ is physical: $\hbar^2 l(l+1)$ is the magnitude squared of the angular momentum. Within each group, rotating the solution gives you a new solution in the same group. These two facts are of course connected: a rotation can't change the length of a vector.
One can show that each group contains $2l+1$ independent solutions, in that any solution $|n,l\rangle$ where the energy is $E_n$ and the angular momentum $\hbar^2 l(l+1)$ can be written as a sum $$|n,l\rangle = \sum_{m=-l}^l c_m |n,l,m\rangle$$ (I apologize for the somewhat poor notation.)
This decomposition is based on choosing a particular axis, and taking each state to depend on the angle $\varphi$ around this axis as $e^{im\varphi}$. The appearance of axes of symmetry in these plots is due to this choice of axis and particular decomposition. With another choice of axis, which is the same as a rotation, the states will be mixed.
The bottom line is that it's not each solution -- wavefunction -- that needs to be spherically symmetric, but the total set of solutions.