As in my comments,
not all subgroups of a direct product is a direct product of subgroups of each factor (although the image of the projection onto a factor is a subgroup of that factor).
The correct way to do this is to use Goursat's lemma:
The set of all subgroups of $G\times H$ is in bijection with
the set of $5$-tuples $(G_1,G_2,H_1,H_2,\varphi)$,
where $G_2\unlhd G_1\le G$, $H_2\unlhd H_1\le H$, and $\varphi$ is an isomorphism from $G_1/G_2$ to $H_1/H_2$.
Any such tuple corresponds to the subgroup
$S=\{(g,h)\in G_1\times H_1: \varphi(gG_2)=hH_2\}$.
For details, I suggest reading a nice answer by @Tobias Kildetoft.
Fix a tuple $(G_1,G_2,H_1,H_2,\varphi)$,
and let $m=|G_1/G_2|=|H_1/H_2|$.
It is not hard to see that $|S|=m|G_2||H_2|$.
Note that $m\in\{1,2\}$ because $m$ divides $\gcd(2\cdot 3,2^4\cdot 17)$.
If $m=1$, Goursat's lemma gives subgroups of the form $G_1\times H_1$,
which you have already obtained:
$$
\begin{array}{c|cc}
\text{Index} & \text{Subgroups}\ (m=1)\\
\hline
2 & \langle \sigma^2 \rangle \times \langle \tau \rangle,\ \langle \sigma \rangle \times \langle \tau^2 \rangle \\
4 & \langle\sigma^2\rangle \times \langle\tau^2\rangle,\ \langle\sigma\rangle \times \langle\tau^4\rangle
\end{array}
$$
Now suppose $m=2$.
To find index $2$ subgroups, we have no choice but choosing
$(G_1,G_2,H_1,H_2)=(\langle\sigma\rangle,\langle\sigma^2\rangle,\langle\tau\rangle,\langle\tau^2\rangle)$.
We have only one choice for $\varphi$ because
the only isomorphism $G_1/G_2\cong\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\cong H_1/H_2$
is the identity.
Applying Goursat's lemma,
we get $S=\langle\sigma^2\rangle\times\langle\tau^2\rangle\ \cup\ \sigma\langle\sigma^2\rangle\times\tau\langle\tau^2\rangle=\langle(\sigma,\tau)\rangle$.
Finding index $4$ subgroups is similar.
The only subgroup of index $4$ is $\langle (\sigma, \tau^2) \rangle$.
We have the following subgroup lattice of the Galois group, including only subgroups of index $1,2,4$:
Now we try to determine some generators for the corresponding fixed fields.$\newcommand{\rat}{\mathbb{Q}}\newcommand{\zt}{\zeta_{17}}\newcommand{\zs}{\zeta_{7}}$
As given by @Jyrki Lahtonen, $\sqrt{17}\in\rat(\zt)$ and $i\sqrt{7}\in\rat(\zs)$.
More explicitly,
$$
\begin{align*}
\sqrt{17}&=\sum_{k=1}^{16} \left( \frac{k}{17} \right) \zt^k \\
i\sqrt{7}&=\sum_{k=1}^{6} \left( \frac{k}{7} \right) \zs^k \\
\end{align*}
$$
where $(\frac{k}{p})$ is the Legendre symbol.$\newcommand{\l}{\langle}\newcommand{\r}{\rangle}\newcommand{\s}{\sigma}\newcommand{\t}{\tau}$
The nonzero quadratic residues modulo a prime $p$ form a subgroup of $(\mathbb{Z}/p\mathbb{Z})^*$.
Using this fact, we see that $\sqrt{17}$ is fixed by $\s$ and $\t^2$ but not $\t$, because $3$ is not a quadratic residue modulo $17$ but $9$ is.
Similarly, $i\sqrt{7}$ is fixed by $\t$ and $\s^2$ but not $\s$.
Finally, $i\sqrt{7}\sqrt{17}$ has degree $2$ over $\rat$
and is fixed by $\sigma\tau$ (using the fact that $\s(i\sqrt{7})=-i\sqrt{7}$ and $\t(\sqrt{17})=-\sqrt{17}$).
We have found generators for all quadratic subfields.
From the subgroup lattice,
we see that only $\frac{\l\s\r\times\l\t\r}{\l\s\r\times\l\t^4\r}$ and $\frac{\l\s\r\times\l\t\r}{\l(\s,\t^2)\r}$ are isomorphic to $\mathbb{Z}/4\mathbb{Z}$.
The number $\gamma=\zt+\zt^{-1}+\zt^4+\zt^{-4}$ is fixed by $\t^4$
but not $\t^2$.
Let
$$
\begin{align*}
\alpha &= \zs-\zs^3+\zs^2-\zs^{-1}+\zs^{-3}-\zs^{-2} = i\sqrt{7} \\
\beta &= \zt-\zt^{-8}+\zt^{-4}-\zt^{-2}+\zt^{-1}-\zt^{8}+\zt^{4}-\zt^{2}
\end{align*}
$$
We have $\alpha\neq 0$, $\beta\neq 0$, and
$$
\s(\alpha)=-\alpha,\ \t^2(\beta)=-\beta
$$
so $\alpha\beta$ is fixed by $\s\t^2$ (but not $\s$, for example).
We have thus found generators for all quartic subfields whose Galois group is isomorphic to $\mathbb{Z}/4\mathbb{Z}$.
Subfield lattice of $\mathbb{Q}(\zeta_{2023})$, including only subfields with degree $1,2,4$:
As a sidenote,
apparently $\beta=\sqrt{\frac{17-\sqrt{17}}{2}}$.
Best Answer
Why did you choose the generators of $(\mathbf Z/16\mathbf Z)^\times$, regarded as $C_2 \times C_4$, so that $(1,0)$ is $\zeta \mapsto \zeta^7$ and $(0,1)$ is $\zeta \mapsto \zeta^3$?
The nicest cyclic decomposition of $(\mathbf Z/16\mathbf Z)^\times$ is $\langle -1 \bmod 16\rangle \times \langle 5 \bmod 16 \rangle$, with the first subgroup being cyclic of order $2$ and the second subgroup being cyclic of order $4$. Thus it seems most natural to let $(1,0)$ correspond to $\zeta \mapsto \zeta^{-1}$ and $(0,1)$ correspond to $\zeta \mapsto \zeta^5$.
Your upside-down subgroup diagram for $C_2 \times C_4$ has mistakes. As an example, the cyclic subgroup $\langle (0,1)\rangle$ with order $4$ has just one subgroup of order $2$, but you wrote two subgroups of $\langle(0,1)\rangle$ with order $2$. As another example, $\langle (1,1)\rangle$ is cyclic of order $4$, so it too has just one subgroup of order $2$, but you gave $\langle (1,1)\rangle$ two subgroups of order $2$.