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
There are a number of different ways to pose this connection. One way is to view the Fourier transform of a periodic function as a distribution, in which case it is a sum of Dirac deltas with weights corresponding to the coefficients in the Fourier series. In this sense the Fourier transform is really the fundamental object.
Another way is to realize the Fourier transform as a limit of Fourier series. Specifically, you can take a function and cut it off outside some large interval. Then you can take the Fourier series of the new function on the interval. Note that there are some questions about how this Fourier series converges if the new function is not periodic. But we are at least guaranteed convergence in $L^2$ if the original function was $L^2$.
Now when you take the size of the interval to infinity, these Fourier series converge in a sense to the Fourier transform. What is going on here is that as the space scale you are considering gets larger, the frequency gap between the relevant Fourier modes is getting smaller. (This is a variant of the uncertainty principle.) As the space scale gets very large, the frequency gap becomes very small, until in the limit there is a continuum of frequencies.