The Dirichlet-Dini Theorem states that the Fourier series for a periodic integrable $f$ converges to $L$ at $\theta$ if the following improper integral exists for some $\delta > 0$.
$$
\int_{0}^{\delta}\frac{1}{\theta'}\left|\frac{f(\theta+\theta')+f(\theta-\theta')}{2}-L\right|d\theta' < \infty.
$$
You don't need conditions elsewhere on the interval except that $f$ is integrable on $[0,2\pi]$. This is a fairly weak condition, but still not weak enough to be necessary.
For example, if $f$ has left- and right-hand derivatives at $\theta$, then the above holds. Or if $f$ has left- and right-hand limits $f(\theta\pm 0)$ and $f$ is $\alpha$-Holder continuous from the left- and the right- for some $0 < \alpha \le 1$. A variety of other conditions will work. The strangest part is that $f$ can blow up near $\theta$ and the series will converge to $0$ if $f$ is symmetric about $\theta$--that's also covered by the above condition with $L=0$. Any of these results covers the case of the step function.
The sufficiency of the Dirchlet-Dini condition can be directly verified from the Dirichlet integral formula for the truncated Fourier series. Non-trivial necessary conditions are not known.
The most impressive, difficult-to-prove result of Fourier Series convergence is Carleson's Theorem: If $f$ is square integrable on $[0,2\pi]$, then the Fourier series converges pointwise a.e. to $f$.
A Fourier series is only defined for functions defined on an interval of finite length, including periodic signals, as you can see from the definition of the Fourier coefficients (in the basis $\{e^{inx}\}_{n\in\mathbb{Z}}$)
$$
a_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}~dx.
$$
You can't define an aperiodic signal on an interval of finite length (if you try, you'll lose information about the signal), so one must use the Fourier transform for such a signal.
In principle, one could take a Fourier transform of a periodic signal in the sense that one could extend the signal outside the interval $[-\pi,\pi]$ by zero. But the resulting Fourier transform would look like
$$
\widehat{f}(p) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-ipx}~dx
$$
and this isn't particularly different from the Fourier coefficients. Moreover, being able to express a periodic signal as a discrete sum of frequencies is a stronger statement than expressing it as a continuous sum via the inversion formula.
Edit: In response to the comment.
I mean "stronger" rather loosely, not in the mathematical sense of one statement implying another.
Taking the Fourier transform of a periodic signal by extending it to $0$ outside $[-\pi,\pi]$ gives no information that the usual Fourier transform would not. Furthermore, the inversion formula still holds, and therefore we see that $f$ can be recovered from its Fourier transform as a continuous sum over all frequencies.
What is remarkable about periodic functions is that one does not actually need all this information to recover the function; out of uncountably many values $\widehat{f}(p)$, one only needs the integer values $\widehat{f}(n)$ (i.e. the Fourier coefficients) to reconstruct the function.
So in this sense, for a periodic function, there is more to say than what the Fourier transform alone provides.
Best Answer
The ("formal") Fourier series $\sum_k\ c_k\ e^{ikt}$ of a given $2\pi$-periodic function $t\mapsto x(t)$ "exists" as soon as $x(\cdot)$ is integrable over an interval of length $2\pi$, so that the coefficients $c_k$ can be computed.
The difficult question is: For which $t$ this series converges to the given function value $x(t)\ $? The series may be useful for the study of $x(\cdot)$ even if for some $t$ it does not converge to the desired value (as, e.g., in the case of a square wave).
A sufficient condition for $\sum_k\ c_k\ e^{ikt}=x(t)$ for all $t$ is the following: The given function should be continuous and of bounded variation on ${\mathbb R}/(2\pi)$. (A square wave is not continuous.) Such a function may very well (but usuallly doesn't) have infinitely many proper local maxima and minima, let alone be constant on subintervals. Consider, e.g., the following function, extended periodically to all of ${\mathbb R}$:
$$x(t)\ :=\ t^2\ \cos{1\over t}\quad\bigl(0<|t|\leq\pi\bigr), \qquad x(0):=0\ .$$