The generalised version of the theorem I'm working with is:
If $R$ is a commutative Ring with unity and $I_1,\ldots , I_n$ ideals of $R$, then the map $$\phi: R \to \frac{R}{I_1}\times\ldots\times \frac{R}{I_n} \ , \ r\mapsto \left(r+I_1, \ldots , r+I_n\right)$$ with $r+I_k$ being the congruence class of $r$ mod $I_k$, is a ring homomorphism with $\ker=\bigcap_{k=1}^{n}I_k$. If in addition the ideals are pairwise comaximal, then we have $$\frac{R}{\bigcap_{k=1}^{n}I_k} \cong \frac{R}{I_1}\times\ldots\times \frac{R}{I_n}$$ I have already managed to prove this, but my problem is to understand how the generalised version relates to the classical one regarding congruence relation of the integers (modular arithmetic). For example, how would I understand a solution $x \in \mathbb{Z}$ to a set of congruence relations (classical version) in terms of the above isomorphism? Where would I find this $x$?
So on the left-hand side of the "sophisticated" version you have $R$ modulo a single ideal (the intersection of all the given ideals, in the integer version a single congruence to a composite modulus divisible by all the given moduli). In this ring is an $x$ which has a counterpart on the right-hand side through the isomorphism.
Now in the integer version you are given congruences to different moduli satisfying the pairwise comaximal condition. The solution to the $r^{th}$ congruence is an element of $\frac R{I_r}$. Combining the solutions gives an element of the right-hand side by specifying all the components. The isomorphism says there is a unique element $x$ on the left-hand side which corresponds to the individual solutions - a single solution of all the equaltions (in the integer version) to the relevant composite modulus.
There are proofs of this for integers which show how to construct a solution. Here the proof simply says that $x$ exists without giving a construction.