There are lots of posts on MSE and the web titled "converse to CRT" but this is not the same. The following is from "Multiplicative number theory I: Classical theory" by Hugh L. Montgomery, Robert C. Vaughan:
In Chinese Remainder Theorem direction is to go from two numbers moduli $q_1$ and $q_2$ to a unique number modulus $q_1q_2$. But this claim of the book is a kind of converse to that. Also it is not of the form $a_1q_2m_1+a_2q_1m_2$ ($m_i$ being reciprocal to $a_i$). I search Internet and worked on it a lot but I can't reach a satisfactory proof for the claim. Anyone knows how $a \mod q_1q_2$ can be written uniquely as $a_1q_2+a_2q_1$?
First, this is only true because $q_1$ and $q_2$ are coprime $\iff (q_1,q_2)=1$: Suppose $\color{red}a \equiv \color{blue}{a_1}q_2+\color{green}{a_2}q_1\equiv \color{blue}{a_3}q_2+\color{green}{a_4}q_1 \pmod{\color{red}{q_1q_2}}$
Then, $\color{red}a \equiv \color{blue}{a_1}q_2\equiv \color{blue}{a_3}q_2 \pmod{\color{red}{q_1}}$ Due to the coprime condition, we have a unique inverse of $q_2$ modulo $q_1$. In other words, $\color{blue}{a_1} \equiv \color{blue}{a_3} \pmod {q_1}$. But since $1\le a_1,a_3\le q_1$, they are equal. Similar argument for $a_2$ and $a_4$.