Quantum Fourier Transform question regarding measurement
In fact the QFT and the DFT are very different. In the DFT we have $N$ complex values stored in several bits per value, which could be called the standard coefficients of this vector in $\mathbb C^N$, and the Fourier transform produces the Fourier coefficients, which are the coefficients of the same vector in a different basis. In this case the coefficients are stored with finite precision and in general precision will be lost in the change of basis.

The QFT is not the same algorithm implemented with quantum gates; in fact the whole encoding of the state changes. The coefficients are not encoded in qubits: if $N = 2^n$ we now need only $n$ qubits to hold the data, and the coefficients of our element of $\mathbb C^N$ are encoded as the coefficients of the general entangled state of these $n$ qubits together (so we have the restriction that they have to sum 1). In theory no precision is lost in the transform. 

As you already remarked, the Fourier coefficients encoded in this way are not accessible to us, and indeed I don't see either how it would be useful to compute a product (I assume you meant product rather than sum) of two polynomials (I am not saying it is not useful for that, just that I don't see how).

If you want to make use of the QFT, it has to be in applications where you are not necessarily interested in the values of all coefficients; e.g. in Shor's algorithm for factoring (or rather period finding) you start out with a (long) periodic sequence of coefficients, and its Fourier transform (possibly after collapsing it to a vector that is 0 everywhere except at some regularly spaced coefficients) will have most coefficients close to 0 but some that are high (and exactly 0 when the period is a divisor of the number of coefficients). You cannot read out the values of the coefficients, but if you look at the qubits (i.e. you measure their values), you will most likely see one of the high coefficients. That gives you a lot of information about the period.

I hope this is (related to) what you  were asking.