Can someone explain tame and wild ramification in cubic integer rings?

You cannot conclude so easily that $2$ and $3$ are ramified. Write $K=\mathbb{Q}(\sqrt[3]{12})$ and $R=\mathbb{Z}[\sqrt[3]{12}]$. Although $2$ and $3$ clearly divide the discriminant $\Delta(R)=\Delta(X^3-12)=-972=-2^4\cdot 3^5$, the relation

$$ \Delta(R)=[\mathcal{O}_K:R]^2\cdot\Delta(\mathcal{O}_K) $$ together with the observation that $2|[\mathcal{O}_K:R]$ if and only if $R$ is singular over $2$, shows that $2$ doesn't divide $\Delta(\mathcal{O}_K)$ if it happens to be the case that $4|[\mathcal{O}_K:R]$. You can prove that this does not happen and that in fact $[\mathcal{O}_K:R]=2$, but this is not immediately clear. However, since $\mathrm{ord}_3(\Delta(R))$ is odd the relation implies immediately that $3|\Delta(\mathcal{O}_K)$, so that is easier.

That aside, you can use some indirect reasoning to determine how $2$ and $3$ factor in $\mathcal{O}_K$. Note that since $R$ is singular over $2$ you don't have factorisation of $2$ in $R$ and you really need to pass to its integral closure $\mathcal{O}_K$.

For example, using the fact that $[\mathcal{O}_K:R]=2$ we see that $\mathrm{ord}_3(\Delta(\mathcal{O}_K))=5$. Now since $3$ ramifies in $\mathcal{O}_K$ we have either $3\mathcal{O}_K=\mathfrak{p}^3$ for a prime $\mathfrak{p}$ of norm $3$ or $3\mathcal{O}_K=\mathfrak{p}^2\mathfrak{q}$ for primes $\mathfrak{p},\mathfrak{q}$ both of norm $3$. If the second case would hold then we have tame ramificiation of $\mathfrak{p}$, and we see that

$$ \mathrm{ord}_{\mathfrak{p}}(\mathfrak{D}_K)=e(\mathfrak{p}/3)-1=1. $$ As $\mathfrak{p}$ is also the only ramified prime over $3$ and the different $\mathfrak{D}_K$ has ideal norm $|\Delta_K|$, we are forced to conclude that $\mathrm{ord}_3(\Delta_K)=1$, which is a contradiction. It follows that $3$ is totally ramified in $\mathcal{O}_K$, and we see that we have wild ramification over $3$.

This technique unfortunately doesn't work for $2$, but it does for any prime $p\neq 2,3$. You'll see that such a $p$ is totally ramified if and only if $\mathrm{ord}_p(\Delta(\mathcal{O}_K))=2$.

To deal with $2$ you use Galois theory or factor $X^3-12$ over $\mathbb{Q}_2$. I'll illustrate the first. Since $\Delta(X^3-12)$ is not a square in $\mathbb{Q}$ it follows that the splitting field $N$ of $X^3-12$, i.e. the normal closure of $K$, is of degree $6$ over $\mathbb{Q}$ with Galois group $S_3$. Since $\Delta(X^3-12)$ is a square in $N$ but not in $\mathbb{Q}$ we obtain the unique quadratic subfield of $N$, which we know must exist by the Galois correspondence. It is $\mathbb{Q}(\sqrt{-2^4\cdot 3^5})=\mathbb{Q}(\sqrt{-3})$.

Now $2$ is unramified (in fact inert) in $\mathbb{Q}(\sqrt{-3})$, so $N/\mathbb{Q}(\sqrt{-3})$ must ramify over $2$, for else $N/\mathbb{Q}$ would be unramified over $2$. Now since $N/\mathbb{Q}(\sqrt{-3})$ is a Galois extenion of prime degree, we must have total ramification over $2$ in $N/\mathbb{Q}(\sqrt{-3})$; this is a consequence of the ram-rel identity $efg=3$ for the extension $N/\mathbb{Q}(\sqrt{-3})$. But then for the ram-rel relation $e_2f_2g_2=6$ in $N/\mathbb{Q}$ we see $2|f_2$ and $3|e_2$, so that $(e_2,f_2,g_2)=(3,2,1)$. From multiplicativity of the ramification index in the tower $N/K/\mathbb{Q}$ it now follows easily that $2$ is totally ramified in $K$, hence we have tame ramification over $2$.

Hope this helps!

EDIT. Kummer-Dedekinds theorem gives an explicit description of the primes of $K$ when $\mathcal{O}_K$ is monogenic, i.e. of the form $\mathbb{Z}[x]$ for some $x\in\mathcal{O}_K$. If this is not the case then finding explicit generators for the primes can be harder (or I'm simply not aware of the techniques).

To try to find such an $x$ in this case, we find a $\mathbb{Z}$-basis for $\mathcal{O}_K$. It is not hard to see that $\{1,\alpha,\beta\}$, with $\alpha=\sqrt[3]{12}$ and $\beta=\alpha^2/2$ is such a basis. If now $x=a+b\alpha+c\beta$ is any element of $\mathcal{O}_K$, we have a $\mathbb{Z}$-linear map $T:\mathcal{O}_K\to\mathcal{O}_K$, which sends $1\mapsto 1$, $\alpha\mapsto x$ and $\beta\mapsto x^2$. Then $\mathrm{im}(T)=\mathbb{Z}[x]$, and $T$ is an isomorphism precisely when $\det(T)=\pm 1$. We try to find $a,b,c\in\mathbb{Z}$ such that this is the case. For this we need to expand $x^2$ in terms of the basis $\{1,\alpha,\beta\}$. From the relations $\alpha^2=2\beta$, $\beta^2=3\alpha$ and $\alpha\beta=6$ we deduce $$ x^2=a^2+b^2\alpha^2+c^2\beta^2+2ab\alpha+2ac\beta+2bc\alpha\beta\\ =a^2+12bc+(2ab+3c^2)\alpha+2(ac+b^2)\beta $$ Thus the matrix representing $T$ with respect to the basis $\{1,\alpha,\beta\}$ takes the form $$ \begin{pmatrix} 1 & a & a^2+12bc \\ 0 & b & 2ab+3c^2 \\ 0 & c & 2(ac+b^2) \end{pmatrix}, $$ which has determinant $2b(ac+b^2)-c(2ab+3c^2)=2b^3-3c^3$. This equals $-1$ for $(a,b,c)=(0,1,1)$, and hence $\mathcal{O}_K=\mathbb{Z}[\alpha+\beta]$. The minimal polynomial of $x=\alpha+\beta$ equals $X^3-18X-30$. Using Kummer-Dedekind one now quickly finds that $2\mathcal{O}_K=\mathfrak{p}_2^3$ for $\mathfrak{p}_2=(2,\alpha+\beta)$, and that $3\mathcal{O}_K=\mathfrak{p}_3^3$ for $\mathfrak{p}_3=(3,\alpha+\beta)$.

It is a simple and well known fact that if $f$ is an Eisenstein polynomial for the prime number $p$, then $p$ is completely ramified in the extension generated by a root of $f$. Clearly $f(x) = x^3 - 12$ is Eisenstein for $3$, and since ${\mathbb Q}(\sqrt[3]{12}) = {\mathbb Q}(\sqrt[3]{18})$ and $g(x) = x^3-18$ is Eisenstein for $p = 2$, the prime $2$ is also completely ramified.

According to pari, the generators of the prime ideals are $2 + \sqrt[3]{12} + \sqrt[3]{18}$ for the prime ideal above $2$ and $3 + \sqrt[3]{12} + \sqrt[3]{18}$ for the prime ideal above $3$. This can be verified by hand in principle: cubing these elements must yield, up to a small power of the fundamental unit $\varepsilon = 1 + 3\sqrt[3]{12} - 3\sqrt[3]{18}$, the elements $2$ and $3$.

In fact we find $(2 + \sqrt[3]{12} + \sqrt[3]{18})^3 = 2(55 + 24\sqrt[3]{12} + 21\sqrt[3]{18}) = 2/\varepsilon$ and $(3 + \sqrt[3]{12} + \sqrt[3]{18})^3 = 3/\varepsilon$.