I'll write $G_n$ for $\mathrm{Gal}(\mathbf Q(\zeta_n)/\mathbf Q)$. You seem
to think that $G_n$ is $\mathbf Z/n\mathbf Z$, where $\zeta_n$ is a primitive
$n$th root of unity; it's not, it's actually $(\mathbf Z/n\mathbf
Z)^\times$. You can use the fact that $\mathbf Q(\zeta_{2n})=\mathbf Q(\zeta_n)$ when $n$ is odd to reduce to the case where $n$ is square-free and odd.
I think you can make something work in a way similar to the proof that the ring
of integers of $\mathbf Q(\zeta_n)$ is $\mathbf Z[\zeta_n]$: start with the case
where $n$ is the power of a prime. Here, $n$ is square-free, so let's look at
the case where $n$ is an odd prime $p$. The conjugates of $\zeta_p$ are
$\zeta_p,\zeta_p^2,\ldots,\zeta_p^{p-1}$, and they clearly constitute a $\mathbf
Z$-basis of $\mathbf Z[\zeta_p]$, so what you said works.
Now if $n$ is odd and square-free, write $n=p_1\cdots p_k$; since the
discriminant of $\mathbf Q(\zeta_{p_i})$ is a power of $p_i$, the discriminants
are pairwise coprime, and so
$$
\mathbf Q(\zeta_n) = \mathbf Q(\zeta_{p_1})\cdots\mathbf Q(\zeta_{p_k}),\\
\mathbf Z[\zeta_n] = \mathbf Z[\zeta_{p_1}]\otimes\cdots\otimes\mathbf
Z[\zeta_{p_k}],\\
G_n = G_{p_1}\times\cdots\times G_{p_k}.
$$
So by induction, all you need is the fact that if $r$ and $s$ are coprime, and
$\zeta_r$ and $\zeta_s$ generate an integral normal basis for $\mathbf
Z[\zeta_r]$ and $\mathbf Z[\zeta_s]$ respectively, then $\zeta_r\zeta_s$ (which
is a primitive $rs$-th root of unity) generates an integral normal basis for
$\mathbf Z[\zeta_r\zeta_s]=\mathbf Z[\zeta_r]\otimes\mathbf Z[\zeta_s]$. Since
you have an isomorphism $G_r\times G_s\simeq G_{rs}$, you can conclude as
follows:
$$ \sum_{\sigma\in G_r}\mathbf Z\sigma(\zeta_r)
\otimes \sum_{\sigma'\in G_s}\mathbf Z\sigma'(\zeta_s)
= \sum_{\tau\in G_{rs}} \mathbf Z\tau(\zeta_r\zeta_s).
$$
I'll write $G=\mathrm{Gal}(\mathbf Q(\theta)/\mathbf Q)$ and
$H=\mathrm{Gal}(\mathbf Q(\theta)/K)$. Take an element $x\in\mathcal O_K$. Then
$x\in\mathcal O_{\mathbf Q(\theta)}$, so there exist integers $m_\sigma$ such
that $x=\sum_{\sigma\in G}m_\sigma\sigma(\theta)$. Now since $x$ is in $\mathcal
O_K$, you have $\tau(x)=x$ for all $\tau\in H$; in other words,
\begin{align}
\sum_{\sigma\in G}m_\sigma\sigma(\theta)
&= \sum_{\sigma\in G}m_\sigma\tau\sigma(\theta) \\
&= \sum_{\sigma\in G}m_{\tau^{-1}\sigma}\sigma(\theta)
\end{align}
by the change of variables $\sigma\mapsto\tau^{-1}\sigma$. Since $\theta$
generates $\mathbf Q(\theta)$, you have
$$ \sum_{\sigma\in G}m_\sigma\sigma
= \sum_{\sigma\in G}m_{\tau^{-1}\sigma}\sigma $$
and Dedekind's lemma on the independence of automorphisms shows that
$m_\sigma=m_{\tau^{-1}\sigma}$ for all $\sigma\in G$ and $\tau\in H$.
Now the idea is to group the terms in $x=\sum m_\sigma\sigma(\tau)$ into
equivalence classes modulo $H$. So let $\Gamma\subset G$ be a lift of $G/H$;
then
\begin{align} x
&= \sum_{\sigma\in G}m_\sigma\sigma(\theta) \\
&= \sum_{\sigma\in\Gamma}\sum_{\tau\in H}m_{\sigma\tau}(\sigma\tau)(\theta) \\
&= \sum_{\sigma\in\Gamma}\sigma\left(\sum_{\tau\in
H}m_{\sigma\tau}\tau(\theta)\right).
\end{align}
I'm going to assume that $K/\mathbf Q$ is Galois, otherwise I don't see how to
conclude. If it is Galois, then $H$ is normal in $G$, so any $\sigma\tau$ can be
written as a $\tau'^{-1}\sigma$, hence $m_{\sigma\tau} = m_{\tau'^{-1}\sigma} =
m_\sigma$, so
\begin{align} x
&= \sum_{\sigma\in\Gamma}m_\sigma\sigma\left(\sum_{\tau\in
H}\tau(\theta)\right) \\
&= \sum_{\sigma\in\Gamma}
m_\sigma\sigma\left(\mathrm{Tr}_{\mathbf Q(\theta)/K}(\theta)\right).
\end{align}
Notice that $\mathrm{Gal}(K/\mathbf Q) = G/H$ and you're done.
