You've gotten several good answers. Let me address the questions in your third paragraph.
I believe the result you are looking for is the following:
Definition. Let $G$ be a group, and let $N$ and $M$ be normal subgroups of $G$. We say that $G$ is the internal direct product of $N$ and $M$ if and only if:
- $N\cap M = \{e\}$; and
- $G=NM$.
And then we have:
Theorem. Let $G$, $H$, and $K$ be groups. The following are equivalent:
- $G$ is isomorphic to the external direct product of $H$ and $K$, $G\cong H\times K$.
- There exist normal subgroups $N$ and $M$ of $G$ such that $N\cong H$, $M\cong K$, and $G$ is the internal direct product of $N$ and $M$.
(There is no requirement that $G$ be abelian).
Proof. To show that (1) implies (2), let $\varphi\colon H\times K \to G$ be the isomorphism. Then let $N=\varphi(H\times\{e\})$, $M=\varphi(\{e\}\times K)$; since $\varphi$ is an isomorphism, and $H\times K = (H\times\{e\})(\{e\}\times K)$ and both subgroups are normal, condition (2) follows.
Conversely, let $\phi\colon N\to H$ and $\psi\colon M\to K$ be isomorphisms, and assume that $G$ is the internal direct product of $N$ and $M$. Define a map $f\colon G\to H\times K$ as follows: given any $g\in G$, then since $G=NM$ we have $g=nm$ for some $n\in N$ and $m\in M$; define $f(g) = (\phi(n),\psi(m))$.
First, note that this is well-defined: if nm=n'm' for n,n'\in N and m,m'\in M, then (n')^{-1}n = m'm^{-1}; but this element is in both $N$ and $M$, and since $N\cap M=\{e\}$, it follows that m'm^{-1}=(n')^{-1}n = e, hence m=m' and n=n'. Thus, the expression $g=nm$ is unique, and since $\phi(n)\in H$ and $\psi(m)\in K$, we have that $f(g)$ is well defined and lies in $H\times K$.
Next, note that if $n,\in N$ and $m\in M$, then the fact that $N\cap M=\{e\}$ implies that $nm=mn$: because $n^{-1}m^{-1}nm = \left( n^{-1}m^{-1}n\right)m = n^{-1}\left(m^{-1}nm\right),$ the middle expression is in $M$, the third expression in $M$; so $n^{-1}m^{-1}nm\in N\cap M = \{e\}$, hence $n^{-1}m^{-1}nm = e$, so multiplying on the left by $mn$ we get $nm=mn$, as claimed.
Thus, to show $f$ is a group homomorphism, let g,g'\in G; write $g=nm$ and g'=n'm'. Then gg' = (nm)(n'm') = n(mn')m' = n(n'm)m' = (nn')(mm'). Hence, f(gg') = (\phi(nn'),\psi(mm')) = (\phi(n)\phi(n'),\psi(m)\psi(m')) = (\phi(n),\psi(m))(\phi(n'),\psi(m')) = f(g)f(g').
Next, to show $f$ is one-to-one, note that if $f(g)=(e,e)$, where $g=nm$, then $\phi(n)=e$ so $n=e$ (since $\phi$ is an isomorphism), and $\phi(m)=e$ so $m=e$. Thus, $g=nm=ee=e$. So $\mathrm{ker}(f)=\{e\}$, hence $f$ is one-to-one.
Finally, to show that $f$ is onto, let $h\in H$ and $k\in K$. Since $\phi$ and $\psi$ are isomorphisms, there exist $n\in N$ and $m\in M$ such that $\phi(n)=h$ and $\psi(m)=k$. Then $f(nm) = (\phi(n),\psi(m)) = (h,k)$. So $f$ is onto. Thus, $f$ is an isomorphism, and we are done. $\Box$
So, in order to show that $D_{12} \cong D_6\times C_2$, you could find normal subgroups $N$ and $M$ of $D_{12}$ such that $N\cong D_6$, $M\cong C_2$, $N\cap M=\{e\}$, and $D_{12}=NM$.
You don't tell us how you think about $D_{12}$; one can think of it as permutations on the set $\{1,2,3,4,5,6\}$ (number the vertices of the regular hexagon and describe the elements of $D_{12}$ by how the vertices are permuted), or abstractly as a group presented as: $D_{12} = \Bigl\langle r,s\;\Bigm|\; r^6 = s^2 = 1,\quad sr = r^{-1}s\Bigr\rangle$ (or maybe other ways of presenting $D_{12}$; thought I think that is the most common one).
Consider the subgroup $N = \langle r^2,s\rangle$. It is not hard to verify that this subgroup has $6$ elements, and so, being of index $2$, it must be normal in $D_{12}$. Then let $M=\{1,r^3\}$. Since $r^3r = rr^3$ and $r^3s = sr^3$, then $M$ is central, and hence normal (verify that $x^{-1}r^3x = r^3$ for every $x\in D_{12}$ using the observations above). And so $M$ is normal and isomorphic to $C_2$.
Then show that $N\cong D_6$ (which is not hard).
Finally, note that $N\cap M = \{e\}$ and therefore $|NM| = |N||M| = 12 = |G|$; since $N$ and $M$ are normal, $NM$ is a subgroup, and so $NM=G$. Thus, you will get, using the theorem, that $G$ is isomorphic to $D_6\times C_2$.