CRT- adding numbers in two different mod worlds

Question

The proof of the CRT goes as follows:
Given the number $x \epsilon Z_m$, $m=m_1m_2...m_k$ $$M_k = m/m_k$$ construct: $$ x = a_1M_1y_1+a_2M_2y_2+...+a_nM_ny_n$$ where $y_k$ is the particular inverse of $M_k\ mod\ m_k$ $$\Rightarrow x\equiv a_kM_ky_k\equiv a_k(mod\ m_k)$$

What I don't understand is:
how is $x\equiv a_1M_1y_1+a_2M_2y_2+...+a_kM_ky_k$ and this lies in $mod\ m$? Is this because there is some rule in modular arithmetic for adding two numbers in two different mod worlds like: $c \mod d \ + e\ mod f = (c+d)(mod(ef))$? As far as I know, there isn't one like that. And how does the addition of these items all in a different mod world provide the solution for x?

Answer 1

The part 2 of the chinese remainder theorem, which starts off at this page and continues to the next, explains the concept of lcm required to understand the OP's question and the concept why the solution exists in mod m, which is actually only a way of finding the minimum solution.

Answer 2

I think $x$ is calculated in $\mathbf Z$, using representatives of $M_k^{-1}\bmod m_k$. Note that the congruence class of $x$ $\bmod m$ is independent of the choice of the representatives.