How do electrons travel through nodes
From the image above and from what you're asking, I presume you are referring to a particle confined in a certain space region. You can find everything you need on the internet but I wanna clarify a few things about QM:


There is no concept in QM for "path of a particle": position and momentum are incompatible observables, which means that they cannot be measured simultaneously each with zero standard deviation: in fact the product of standard deviations for position and momentum must be greater than $\dfrac{\hbar}{2}$.
$$\sigma_{x}\sigma_{p}\geq\dfrac{\hbar}{2}$$
So it's meaningless in QM to talk about trajectories.
Meaning of the wave function: Let's assume that your system consists of a single particle in a one dimensional space. The wave function $\Psi\left(x,t\right)$
is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared $\left|\Psi\left(x,t\right)\right|^{2}$
, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx
, gives you the probability for finding the particle between $x$
and $x+dx$
$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$
Of course you can integrate the previous eq. to find the probability over a certain region of space $\Omega$
$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$
Stationary state: If the potential is time independent (in your case it's zero), the time dependence of the wave function can be factored out like
$$\Psi\left(x,t\right)=\mathrm{e}^{-i\omega t}\psi\left(x\right)$$
In this case it's easy to prove that the probability density is time independent.
The quantization: For this simple problem, you must solve the eigenvalue equation for the energy
$$-\dfrac{\hbar^{2}}{2m}\dfrac{d\psi\left(x\right)}{dx}=E\psi\left(x\right)$$
By solving it and by imposing that the particle cannot exists outside a certain region (outside the region $\left[0,3\right]$ in your picture), you find a natural quantization condition for the energy: the energy can assume only certain discrete values, which are called eigenvalues, $E_{n}$
. Of course, this means that for every eigenvalue there is an associated wave function, which is called eigenfunction $\psi_{n}\left(x\right)$
. The general solution to the eigenvalue equation is in fact a linear combination of eigenfunctions $$\psi\left(x\right)=\sum_{n}c_{n}\psi_{n}\left(x\right)$$
In the above picture, you can see the first three $\left|\psi_{n}\left(x\right)\right|^{2}$
. They of course represent probability density functions: note that where $\left|\psi_{n}\left(x\right)\right|^{2}$
is zero the probability of finding the particle is of course zero (for example in 0
and 3
, like we requested before).
So where is the electron? Like said before, in QM locating a particle it's a non-sense. All you can do is to compute the average position and momentum, which for a system in the state $\psi\left(x\right)$ are given by
$$\left\langle x\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{x}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)x\psi\left(x\right)dx$$
$$\left\langle p\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{p}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\left[-i\hbar\dfrac{d\psi\left(x\right)}{dx}\right]dx$$