I have seen two motivations of decomposing an $L^2(\mathbb{T}^n)$ function into a Fourier series with basis functions $e^{2\pi inx}$ with $n$ being integers. The first one is the intuitive frequency representation of a signal; the second one is that the trigonometrical functions are eigenfunctions of the Laplacian, so it can help to solve PDE. I am satisfied by those explanations.
However, in Folland's Real Analysis, Modern Techniques and Their Applications, it gives another motivation as follows
I feel a little "unmotivated". Surely, having functions in the form $f(y+x)=\phi(x)f(y)$ can be good. But why is it deterministic for people to consider them as the building blocks? Is there any theorem in Fourier analysis that particularly uses this fact? Sorry for being a vague question, but basically, I feel it is somehow unnatural.
Let $G$ be an abelian topological group. Just as we look at the dual space $X^*$ when we want to know about a vector space (a Banach space, for example) $X$, it is a natural idea that the "dual group" $G^*$ might give some useful information about $G$, and this is indeed the case. More concretely, we will consider the complex homomorphisms (usually called the characters) on $G$, which are continuous functions $\phi:G\to\mathbb{C}$ satisfying $$\phi(x+y)=\phi(x)\phi(y)\qquad(x,y\in G). $$ In our case, $G=\mathbb{R}^n$ with addition (in particular, it is abelian), and the characters are precisely given by $$\phi_t(x)=e^{it\cdot x} $$ for some $t\in\mathbb{R}^d$.
There is (at least) one more motivation to consider the characters. Let $G$ be a locally compact abelian group. Then $G$ admits a unique Haar measure, and there arises a natural Banach space $L^1(G)=L^1(G,m)$. In fact, $L^1(G)$ is an abelian Banach algebra whose multiplication is given by the convolution. Then it is well known that every complex homomorphism on $L^1(G)$, that is, every function $h:L^1(G)\to\mathbb{C}$ such that $h(f*g)=h(f)h(g)$ is given by $$h_{\phi}(f)=\int_Gf(x)\phi(-x)dx\qquad(\phi\in G^*). $$ So again, characters enter naturally. We say that $h$ is the Fourier transform of $f$ at $\phi$ and write $\widehat{f}(\phi)=h_{\phi}(f)$. Note that this coincides with the usual definition in $\mathbb{R}^n$ if we identify $\mathbb{R^n}$ and its dual group by $t \longleftrightarrow \phi_t$.