Basic Confusion About the DFT and Convolution
As I understand your question, it boils down to: why is it ok, to take the convolution of a periodic signal only on an interval rather than over infinity? The DFT or discrete nature of the signal does not really play a role in your question. It merely provides the basis of assumung your two signals $f$ and $g$ as periodic, on the interval $[0,T]$ such that $$f(t+T)=f(t-T)=f(t)$$ and obviously $g(t+T)=g(t-T)=g(t)$.

Now assume that you define the convolution over the whole $(-\infty,\infty)$ like so:
$$y(t):=(f\ast g)(t)=\int_{-\infty}^\infty d\tau\, f(t-\tau)g(\tau)$$

From the periodicity above follows that
$$\int_{-\infty}^\infty d\tau\, f(t-\tau)g(\tau) = \int_{-\infty}^\infty d\tau\, f(t+T-\tau)g(\tau)$$
which you can intepret intuitevely to mean that the convolution should also be periodic. I write "intuitively" because the integral is divergent as we will see below.

Furthermore you can split the integral into a sum of integrals where $\tau$ goes over intervals of length $T$, such as $[n\cdot T,(n+1)\cdot T]$ with $n$ a signed integer number in $(-\infty,\infty)$. Doing that we have (after some simplifications)

$$\begin{eqnarray}
y(t)&amp;=&amp;(f\ast g)(t)=\int_{-\infty}^\infty d\tau\, f(t-\tau)g(\tau) \\
&amp;=&amp;\sum_{n=-\infty}^\infty \int_0^Td\tau\,f(t-\tau)g(\tau)  
\end{eqnarray}$$ 
which would be divergent just because of the sum. Thus, the convolution of periodic functions produces a divergent integral. Sticking to this definition of the convolution would make it impossible to define the convolution like that. At least one of the functions must go to zero sufficiently fast, to produce a finite integral. Your periodic functios, however, will violate this decay condition. Thus it makes sense to window the functions with a rect function that captures all the information in one period.

So you see why it would make sense to define the cyclic convolution as just the integral

$$y(t)=\int_0^Td\tau\,f(t-\tau)g(\tau)$$

and use the periodic extension, since we already provided some intuition as to why setting $y(t+T)=y(t)$ makes sense. Thus the cyclic convolution carries all the information you would have from the regular definition but avoids the divergence due to the infinite sum. I hope this helps.