Let $a,b,c$ be arbitrary integers such that $a$ is odd and $(a,b,c)=1$.
Let $R = \mathbb{Z}_8$, the set of all integer residues modulo $8$. Define an $R$-module endomorphism $\phi \colon R \times R$ by \begin{align*} \phi(x,y) = (2ax+by,y) \end{align*}
We have that $\ker(\phi) = \langle 4 \rangle \times \langle 0 \rangle$, where $\langle g \rangle$ is the principal ideal generated by $g$. By the 1st isomorphism theorem for modules, we have that \begin{align*} \phi(R \times R) \simeq (R \times R)/(\langle 4 \rangle \times \langle 0 \rangle) \end{align*}
One can now show that $\phi(R \times R) \simeq R/\langle 4 \rangle \times R$, see, for example, exercise 11, section 10.2 in Dummit and Foote.
Observe that $R/\langle 4 \rangle$ is a quotient module under the ring action \begin{align*} r(x + \langle 4 \rangle) = rx + \langle 4 \rangle. \end{align*}
One can see that $ R/\langle 4 \rangle \simeq \langle 2 \rangle$ and so we deduce \begin{align*} \phi(R \times R) \simeq \langle 2 \rangle\times R \end{align*}
Set $\displaystyle e(\alpha) = e^{2\pi \imath \alpha}$ for any rational number $\alpha$.
Observe that \begin{align*} 4a(ax^2+bxy+cy^2) = (2ax+by)^2+(4ac-b^2)y^2. \end{align*}
Hence, now consider the sum \begin{align*} \sum_{x,y=0}^{2-1} e\left(\frac{ax^2+bxy+cy^2}{2}\right) &= \left(\frac{1}{2^2}\right)^2 \sum_{x,y=0}^{8-1} e\left(\frac{ax^2+bxy+cy^2}2\right)\\ &=\frac{1}{2^4}\sum_{x,y=0}^{8-1} e\left(\frac{a^{-1}(4a)(ax^2+bxy+cy^2)}{8}\right)\\ &= \frac{1}{2^4}\sum_{y=0}^{8-1} e\left(\frac{a^{-1}(4ac-b^2)y^2}{8}\right) \sum_{x=0}^{8-1} e\left(\frac{a^{-1}(2ax+by)^2}{8}\right). \end{align*}
By the isomorphism above, we can replace $2ax+by$ with $2x$, so that for arbitrary $y$, as $x$ runs over a complete residue system modulo $8$, $2x$ runs over an isomorphic set of residues to $2ax+by$. Hence, this yields
\begin{align*} \frac{1}{2^4}\sum_{y=0}^{8-1} e\left(\frac{a^{-1}(4ac-b^2)y^2}{8}\right) \sum_{x=0}^{8-1} e\left(\frac{a^{-1}x^2}{2}\right)=0 \end{align*}
However, when we look at the sum itself, we have \begin{align*} \sum_{x,y=0}^{1} e\left(\frac{ax^2+bxy+cy^2}2\right) &= 1 + (-1)^a + (-1)^c + (-1)^{a+b+c}\\ &=(-1)^c(1+(-1)^{b-1}). \end{align*}
This last expression is non-zero whenever $b$ is odd.
Thus, I have made a mistake somewhere in my module homomorphisms. I think the obvious mistake is that $2ax+by$ doesn't "behave" like $2x$ if $b$ is odd. But I don't see the error in the isomorphism above. Any help would be great!