Let $X=\mathbb{Z}_{12}^k$ for $k\in \mathbb{N}$ and $G=D_{12}$. Define an action of $D_{12}$ on $X$ by setting rotations $r^n(p)=(p_1+n,\dotsc,p_k+n)$ where the coordinates are taken modulo $12$ and reflections $s(p)=(-p_1,\dotsc,-p_k)$ for $p=(p_1,\dotsc,p_n)\in X$. Then by Burnside's lemma, the number of orbits $Gx=\{gx:g\in G\}$ is given by $|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|$ where $X^g=\{x\in X: gx=x\}$.
So, $$|\mathbb{Z}_{12}^k/D_{12}|=\frac1{24}[12^k+|X^s|+\sum_{j=1}^{11}|X^{r^j}|+\sum_{j=1}^{11}|X^{r^js}|]$$
It is easy to see that $X^s=\{(p_1,\dotsc,p_k):p_i=0,6\}$ hence $|X^s|=2^k$. Now $X^{r_j}=\emptyset$ since $r^j(p)=p$ implies $p_i+j=p_i$ for $i=1,2,\dotsc,k$ which in turn implies $j\equiv_{12} 0$, so $r^j$ only fixes $p$ when it is the identity.
We claim $X^{r^{2j+1}s}=\emptyset$. For $r^{2j+1}s(p)=p$ implies $-p_i+2j+1=p_i$ for $i=1,2,\dotsc,k$ i.e. $2j+1=2p_i$, a contradiction.
Finally, we claim $X^{r^{2j}s}=\{(j,j,\dotsc,j),(j+6,j+6,\dotsc,j+6)\}$. For $r^js(p)=p$ implies $-p_i+j=p_i$ which implies $2p_i-j\equiv_{12} 0$ hence $p_i=j/2+6m$ for some $m$, which only makes sense if $j$ is even. The claim follows.
All together, we would then obtain $$|\mathbb{Z}_{12}^k/D_{12}|=\frac1{24}[12^k+2^k+5\cdot 2]$$ but one can check this fails to make sense for $k=2$, as we get $\frac1{24}[12^2+2^2+10]=\frac{158}{24}$.
So I obviously am missing some fixed points or doing something completely incorrect. If anyone can point out what it is I'd be very grateful! Thanks.
Your treatment of $r^{2j}s$ distinguishes between $j=0$ and $j\ne0$ without reason. The fixed points of $r^{2j}s$ can have $j$ and $j+6$ independently in each coordinate for $j\ne0$ just like for $j=0$. Thus
$$ |\mathbb{Z}_{12}^k/D_{12}|=\frac1{24}\left[12^k+6\cdot2^k\right]\;, $$
and for $k=2$ you get
$$ |\mathbb{Z}_{12}^2/D_{12}|=\frac1{24}\left[12^2+6\cdot2^2\right]=7\;, $$
corresponding to the $7$ inequivalent differences $0,\ldots,6$ between the two entries.