How can this isomorphism be valid? How can $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)> \cong \mathbb{Z}_{12}  = \mathbb{Z}_{4} \times \mathbb{Z}_{3}$?
I am not convinced at the least that $\mathbb{Z}_{12}$ is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$
For instance, doesn't $<1>$ have an order of 12 in $\mathbb{Z}_{12}$? And no element of $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$ can even have an order of $12$ no? 
What is the maximum order of $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(2,3)>$? I know if the denominator isn't there, it is $lcm(4,6) = 12$, but even if it weren't there, I don't see how either component can produce an element of order 12. 
What I mean is that the first component is in $\mathbb{Z}_4$, so all elements have max order 6
and likewise $\mathbb{Z}_6$, have order 6, so how can any element exceed the order of their group?

I presume you mean ring isomorphism. If so, you should define what is the 0, the 1, the sum and the multiplication. Of course, those are rather implicit. But there lies your doubts. As @KReiser points out, the unit in $\mathbb{Z}_4 \times \mathbb{Z}_6 / \langle(2,3)\rangle$ is (the class of) (1,1).
In order to prove the isomorphism to $\mathbb{Z}_{12}$, you can define the bijective function and show that it is an isomorphism of rings. That is, let $\Phi$ be the function that maps $1\in\mathbb{Z}_{12}$ to the class of $(1,1)\in \mathbb{Z}_4 \times \mathbb{Z}_6 / \langle(2,3)\rangle$. First, show that it is a morphism, that it is injective and surjective. Since both sets have the same amount of equivalence classes, they thus turn out to be isomorphic.
