An electromagnetic wave is made of an electric wave and a magnetic wave, so 2 waves.
How many waves are there for a particle wave?
for a photon: if it is not made of two waves, then I don't understand how the photon wave could be compatible with the electromagnetic wave
for an electron: one wave or two waves?
For a particle propagating through space, its wavefunction furnishes the probability of finding it in a given place which is a different number for every different place. This is a single-valued variable.
Electromagnetic waves are very different; they propagate through space as the interplay between a linked pair of an electric field and a magnetic field, and so a wave description of the photon requires the specification of both the electric and the magnetic fields at a single point in space as well as their respective rates of change.
You question begins
An electromagnetic wave is made of an electric wave and a magnetic wave, so 2 waves.
and asks “How many waves for a particle wave?”
However, I think it’s important to realize that “there are two waves” is just one model of the electromagnetic field, and it’s not necessarily the best one. For one thing, electric field $\vec E$ and magnetic field $\vec B$ are both three-dimensional vectors, so you might say more accurately than an electromagnetic wave is six waves: three for $E_x, E_y, E_z$, and three for $B_x, B_y, B_z$. But that’s actually too many degrees of freedom, because $\vec E$ and $\vec B$ are related to each other by Maxwell’s equations. Another way to describe electromagnetic fields and their waves is to introduce the four-potential,
$$ A_\mu = (\phi, \vec A) $$
and the antisymmetric field tensor
$$ F_{\mu\nu} = \partial_\mu A_\mu - \partial_\nu A_\mu $$
Here $\phi$ is the electrostatic potential whose gradient is $\vec E$, and $\vec A$ is the vector potential whose curl is $\vec B$. The six independent and nonzero components of $F_{\mu\nu}$ are the six components of the electric and magnetic vector fields. The Maxwell equations arise when one applies the Euler-Lagrange method to the Lagrangian density
$$ \mathcal L = -\frac1{4\mu_0} F_{\mu\nu}F^{\mu\nu} - J^\mu A_\mu $$
So if there aren’t any charges or currents around ($J^\mu = 0$), you can describe an electromagnetic wave using just the four parameters $A_\mu$. Maybe it’s better to say that an electromagnetic wave is four waves, since $A_\mu$ has four components. And it turns out this is pretty close to the way people describe photon behavior in quantum field theory. The complicated sum of derivatives $F_{\mu\nu} F^{\mu\nu}$ describes the “free field”; the term $J^\mu A_\mu$ is the “interaction term” which couples the “photon field” $A_\mu$ to charges and currents represented by the four-current $J_\mu$.
But it’s not really the case that the oscillations of the $A_\mu$ describe four independent waves, where you can change one without affecting the others. If you do a fair amount of work (more than will fit in this answer) you can show that the four $A_\mu$ can describe two independent waves, which we observe experimentally in electromagnetic waves as “polarization states.” However in the $\vec E, \vec B$ language it’s more parsimonious to talk about polarization in the linear basis, while from the QED direction there is an important connection between the circular polarization basis and the angular momentum of the photon. Furthermore the reason that there are two independent polarizations in electromagnetism, rather than more, arises in a sneaky way from the fact that the photon field $A_\mu$ is massless.
So whether wavelike oscillations in your photon field are described as having two degrees of freedom (in one of the polarization bases) or four (in the $A_\mu$ language) or six (in the $\vec E, \vec B$ language) or perhaps three (because $\vec B$ can be deduced from $\vec E$, if you’re willing to make some assumptions) or just one (because all of the components in the four-potential $A_\mu$ are required to construct an object which is invariant under Lorentz transformations): any of these possibilities could be “correct” under the right circumstances. It’s not a question with a straightforward answer.
You have a similar circumstance talking about matter waves. For an electron you have a complex wavefunction — is an oscillation in complex field a “wave” in one complex number, or in two real ones? Then you discover that you actually have coupled complex probabilities for the two electron spin states. Then you follow Dirac’s argument about Lorentz symmetry and discover that you actually have four coupled waves in the “electron field,” two spin states with negative charge and two with positive charge. The number of degrees of freedom depends on what problem you are trying to solve and how careful you’re being about it.
The answers are good, but I need to explicitly state what is conceptually wrong in your question.
An electromagnetic wave is made of an electric wave and a magnetic wave, so 2 waves.
The classical electromagnetic wave is one wave, its intensity and direction depending on two variables that are a function of each other through the solutions of Maxwell's equations. A simple functional relation of E to B is seen in the plane wave solutions
How many waves are there for a particle wave?
As stated in the other answers, the waves in particle representations are probability waves and depend on the solution of the quantum mechanical equation and boundary conditions of the given interaction.
for a photon: if it is not made of two waves, then I don't understand how the photon wave could be compatible with the electromagnetic wave
The photon is a point particle in the standard model of particle physics and is the quantum of the classical electromagnetic wave. It can be proven that the classical wave is a superposition of a large number of photons, and also this is seen experimentally, see here.
for an electron: one wave or two waves?
There is one wavefunction for each particle depending on the boundary conditions and the interactions, and it gives the probability of observing the particle.
Actually in electro-magnetic wave there is effectively only one "wave", because the electric and magnetic components are strictly coupled. (Not only in waves, but also in general.) Also photons and electro-magnetic waves are the "exact" same things. The names are only distinguished to emhpasise if we are talking about the particle properties, or the wave properties of the same thing.
In particle "waves" there is no physical field that oscillates, but only probabilities. ("In the particle waves a number oscillates.") Thus it is more convenient to talk about a wave function of a particle, rather than a particle wave. Also a particle is usually described by a wave packet, and not by a single (sinusodial plane) wave.
