[Analytic Number Theory - Florian Luca and Jean Marie De Koninck, chapter 14, question 14.5]
A character is called primitive modulo $k$ if it is not induced from any divisor $d<k$, or in other words, if for every divisor $d$ of $k$, there is a number $a\equiv 1\pmod d$ with $(a,k)=1$ and $\chi(a)\neq 1$. Show that if $\chi$ is primitive modulo $k$, then $$\chi(n)=\frac{\tau_k(\chi)}{\sqrt k}\sum_{m=1}^k \bar\chi(m)e^{-\frac{2\pi imn}{k}}$$ where $$\tau_k(\chi)=\frac{G(1,\chi)}{\sqrt k}$$ if $\chi$ is primitive modulo $k$.
In previous exercises, I had to prove that if $$G(n,\chi)=\sum_{m=1}^k \chi(m)e^{\frac{2\pi imn}{k}}$$ then assuming $(n,k)=1$, we have $$G(n,\chi)=\bar\chi(n)G(1,\chi)$$ and $$|G(1,\chi)|^2 = p$$ where $k=p$ is a prime.
These two are enough to solve the problem for the special case $(n,k)=1$. But, I can't figure out how to deal with the case $(n,k)>1$.
Note: The initial version of this post was different due to a typo in the original question. The main problem now is in the case $(n,k)>1$. I am sorry for the confusion.
This is a typo. To produce an easy counterexample, choose your favorite primitive character and compute some random values. Say, for example, we choose the unique primitive character mod $4$, which we write $\chi_{-4}$ as it has Kronecker symbol $(-4/\cdot)$. This is the character that has $\chi_{-4}(1) = 1$ and $\chi_{-4}(-1) = -1$. Note also that as $1 \equiv -1 \bmod 2$ and as $2$ is the only nontrivial divisor of $4$, $\chi_{-4}$ is primitive.
Let's compute $\chi_{-4}(2) = 0$ according to the given formula, $$\chi_{-4}(2)=\frac{\tau_2(\chi_{-4})}{2} \bar\chi(1)e^{-\frac{4\pi i}{4}} = \frac{-1}{2} \tau_2(\chi_{-4}) + 0.$$ This is already enough to stop, as the first term on the right is nonzero (as the Gauss sum has size $2$), and this is a contradiction.
The summation should go up to $k$.