I have heard that the groups $(\mathbb R,+)$ and $(\mathbb Z,+)$ are the most important groups for Fourier series. Why is this the case?
Supposedly, it has something to do with the fact that for any $f$ the set $S:=\{ p \ | \ p \text{ is a period of } f \}$ is a subgroup of $(\mathbb R,+).$
In this answer, I'll give you somewhat of a tour of LCA (locally compact abelian) group theory as it will give you an idea for why these two groups are very nice for Fourier theory. The punchline is that Fourier theory is inherently built around the nice algebraic structures of the reals and integers. If you pay very close attention to proofs regarding Fourier series or the Fourier transform, this becomes very clear.
First note that on the real line we have the Fourier transform given by:
$$\mathcal{F}f(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i\omega t} f(t)\,dt,$$
where $f$ is sufficiently nice and furthermore,
$$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{i\omega t}\mathcal{F}f(\omega)\,d\omega.$$
However, on the unit interval we have Fourier series:
$$f(t) = \sum_{n=-\infty}^{\infty} c_n e^{int},$$
where $c_n = \dfrac{1}{2\pi}\int_0^1 f(t)e^{-int}\,dt$ for sufficiently nice $f$. In our first example, we are summing over real numbers (our integral is over real numbers) whereas in our second example our indexing set---the set we are summing over---is the integers. In one example, we have a continuous sum (integral); in the other, we have a discrete sum. Note that $c_n$ is replaced by $\mathcal{F}f(\omega)$ in the Fourier transform case.
These two phenomena are very, very similar in nature with some very distinct caveats as noted above. Due to their similarity, it seems reasonable to think that there is a deeper theory underlying these two phenomena. To extract this, we need some observations.
Firstly, we had to "sum" over the elements of some set ($\Bbb R$ and $\Bbb Z$). This is the realm of measure theory and suggests that the kinds of structures we can consider must have what are called measures and these measures should be fairly nice. One thing that comes up time and again in Fourier transform theory is that we have to make a change of variable $x'=x+a$. In doing so, we make use of the chain rule to get that $dx' = dx$ so that our integrals are basically translation invariant. Without this, a lot of Fourier theory would be dead in its tracks. Tracing this back to the measures, this means that the measures we are after should also be translation-invariant, i.e. $E$ and $gE$ should have the same volume.
Secondly, many of the proof techniques regarding the Fourier transform and Fourier series require the nice algebraic structures that the reals and integers have so that we can combine the exponentials and manipulate them to our liking. Thus suggests that maybe we need a nice group structure if we are to develop a deeper understanding of Fourier theory.
These two ideas together almost force you into the realm of what are known as locally compact abelian groups. These are abelian groups that have nice topological properties. Such groups admit what is known as a Haar measure, which is a very natural way to codify the volumes of the subsets of the group. Particularly, the Haar measure is translation-invariant. That is, if $E$ is a subset of $G$, then $gE$ and $E$ have the same volume as desired.
A result by Oxtoby shows that if we leave the realm of local compactness (even if we assume our group's topology comes from a metric---usually metric topologies are very strong and guarantee a lot of nice properties), we have no hope of creating a Haar measure - this more or less demolishes any hope of establishing Fourier theory in this setting since we really need our measure (notion of volume) to be translation-invariant in order to say anything meaningful beyond simply defining the Fourier transform on the group.
Whenever measures are present, one can develop a notion of integration on that set. Since we can in fact integrate over locally compact abelian groups, we can potentially devise a Fourier transform on them. However the question becomes: what plays the role of our exponentials? This requires us to consider what exponentials really are algebraically. First, we know that $|e^{ix}| = 1$ if $x$ is real (or a fortiori integral) and also that $e^{i(x+y)} = e^{ix}e^{iy}$. Therefore, exponentials are really nothing more than homomorphisms from $\Bbb R$ and $\Bbb Z$ into the torus (unit circle) $\Bbb T$.
With these pieces together, we can then devise a Fourier transform on any locally compact abelian group. If our LCA group is $G$, then let $\chi:G\to\Bbb T$ be a homomorphism from $G$ into $\Bbb T$. The Fourier transform on the group is nothing more than
$$\mathcal{F}f(\chi) = \int_G \overline{\chi(g)}f(g)\,dg,$$
where $dg$ denotes that we are integrating over our group. In the case of $\Bbb R$, $\chi(x) = \exp(2\pi ixy)$ for some $y\in \Bbb R$ and the integral is our usual notion of integration; in the case of $\Bbb Z$, $\chi(n) = \exp(int)$ for some $t\in[0,1)$ and the Haar measure is the counting measure (which gives us a discrete sum instead of an integral). Furthermore, many of the results of Fourier transform theory transfer over to the general setting of LCA groups, although the proofs are much more arduous.
There are more examples for which this program works than just $\Bbb R$ and $\Bbb Z$, including $\Bbb T$ and what are known as the $p$-adic integers ($\Bbb Q_p$). (I think) Any locally compact abelian group is of the form $\Bbb R^k \times \Bbb Z^l \times \Bbb T^m \times (\Bbb Q_p)^n \times K$ where $K$ is some compact group (could be a finite group like $\Bbb Z/n\Bbb Z$!). Do not quote me on this - I only know this secondhand from a former advisor.
With this classification, it becomes clear that Fourier theory is almost limited to these cases that you know because the algebraic structure is so rich and allows for an obvious generalization.