What is perhaps more useful is to think about how one proves that differentiability/decay implications in the periodic or Fourier transform case. (Think about that for a bit here. Ready?)
Suppose your function is $C^k$. Then we have that $f^{(k)}$ is uniformly bounded. Which means that $a(\lambda, f^{(k)})$ is uniformly bounded. But using that $a(\lambda, f^{(k)}) = (i\lambda)^k a(\lambda,f)$, you see that you have
$$ \sup_{\lambda \in C} |\lambda^k a(\lambda,f)| < \infty $$
follows from $f$ being $C^k$. In general this is the best you can ask for, considering the case that $f = \exp ix $. In the case of the periodic functions you can do one better, in that $ \lim_{n\to \infty} |n^k c_n| = 0$, is because you have the Riemann-Lebesgue lemma. For suitable definitions of almost periodicity, you can get something similar: if you assume your derivative function $f^{(k)}$ is almost periodic in the Besicovich sense, then you will have summability of $\lambda^k a(\lambda,f)$ which will in particular imply that
$$ \lim_{n\to \infty} \sup_{\lambda \in C, |\lambda| > n} |\lambda^k a(\lambda,f)| = 0$$
For the reverse direction, you can get something similar. If you know that $f$ is given as a convergent sum of $\sum a(\lambda,f)e^{i\lambda x}$, you see that immediately, summability of
$$ \sum_{\lambda\in C} |\lambda^k a(\lambda,f)| = S_k < \infty $$
implies that the $k$th derivative of $f$ is uniformly bounded, and that $f$ is at least $C^{k-1}$. Furthermore, if $\sup_{\lambda\in C} |\lambda| < \Lambda < \infty$, you have that your function $f^{(k)}$ is Lipschitz continuous with constant $S_k\Lambda$. So this actually implies that using, in addition, the following
$$ S_{n,k} := \sum_{\lambda \in C, |\lambda| > n} |\lambda^k a(\lambda,f)| $$
with
$$ \lim_{n \to \infty} S_{n,k} = 0 $$
that $f$ is $C^k$. (For $\epsilon$, choose $N$ large enough so that $3S_{N,k}< \epsilon$, then choose $\delta$ such that $3\delta N S_k < \epsilon$. Then do a high-low frequency splitting to show that $|f(x) - f(y)| \leq |x-y| N S_k + 2 S_{N,k}$.) So just a decay condition on the frequency is not enough: you need summability. (In the case of periodic functions, since there is a minimal spacing between frequencies, decay conditions can be directly translated to summability.)
"Are there any other well-known periodic functions?"
In one sense, the answer is "no". Every reasonable periodic complex-valued function $f$ of a real variable can be represented as an infinite linear combination of sines and cosines with periods equal the period $\tau$ of $f$, or equal to $\tau/2$ or to $\tau/3$, etc. See Fourier series.
There are also doubly periodic functions of a complex variable, called elliptic functions. If one restricts one of these to the real axis, one can find a Fourier series, but one doesn't do such restrictions, as far as I know, in studying these functions. See Weierstrass's elliptic functions and Jacobi elliptic functions.
Best Answer
The notation will be simpler with the complex version of Fourier series:
Say $$f(x)=\sum_kc_ke^{ikx}.$$Then for each $k_0$ we have$$|k_0|^{2n}|c_{k_0}|^2 \le\sum_k|k|^{2n}|c_k|^2=||f^{(n)}||_2^2\le||f^{(n)}||_\infty^2\le1.$$
If $|k_0|>1$ then letting $n\to\infty$ shows that $c_{k_0}=0$. So $$f(x)=\sum_{|k|\le1}c_ke^{ikx}=a\cos(x)+b\sin(x)+c.$$Now you just have to figure out what $a,b,c$ actually give $||f^{(n)}||_\infty\le 1$. (Start with some trig, rewriting $a\cos(x)+b\sin(x)=A\cos(x+\phi)$; the condition I get is $(a^2+b^2)^{1/2}+|c|\le1$.)