The correct formula is, for $\chi$ primitive of conductor $q$,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n)\hat f\left(\frac{n/x}{\sqrt{q}}\right). $$
Here $A:=\sqrt{q}/\tau(\bar\chi)$ is the so-called root number, it is of modulus $1$.

This formula is essentially equivalent to the functional equation of $L(s,\chi)$, but let me provide a direct proof. We start from the well-known formula
$$ \chi(n) = \frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m)e\left(\frac{mn}{q}\right), $$
where $e(x)$ abbreviates $e^{2\pi i x}$. See
Davenport: Multiplicative number theory, Chapter IX, equation (6).
Then we can rewrite the left hand side in the formula as
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m) \sum_{n\in\mathbb{Z}}
e\left(\frac{mn}{q}\right) f\left(\frac{nx}{\sqrt{q}}\right).$$
Applying the Poission summation formula for the inner sum on the right hand side,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{1}{\tau(\bar\chi)}\sum_{m=1}^q\bar\chi(m)
\sum_{k\in\mathbb{Z}}\int_{-\infty}^\infty e\left(\frac{mt}{q}+kt\right) f\left(\frac{tx}{\sqrt{q}}\right)\,dt. $$
Denoting $n:=m+qk$ on the right hand side, we obtain
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{1}{\tau(\bar\chi)}\sum_{n\in\mathbb{Z}}\bar\chi(n)
\int_{-\infty}^\infty e\left(\frac{nt}{q}\right) f\left(\frac{tx}{\sqrt{q}}\right)\,dt. $$
By a change of variable,
$$ \sum_{n\in\mathbb{Z}}\chi(n)f\left(\frac{nx}{\sqrt{q}}\right) =
\frac{A}{x}\sum_{n\in\mathbb{Z}}\bar\chi(n)
\int_{-\infty}^\infty e\left(\frac{nt/x}{\sqrt{q}}\right) f(t)\,dt, $$
where $A$ is as above. This is the stated (corrected) formula.

**Remark of 10/14/2013.** The formula as proved above holds for integrable continuous functions satisfying $|f(t)|+|\hat f(t)|\ll (1+|t|)^{-1-\delta}$ for some $\delta>0$. I had an Addendum of 10/12/2013 here with a seemingly more relaxed condition for $q>1$, but this condition turned out to be equivalent to the original one by basic facts on the Fourier transform.