Let $R$ be a PID and $$\varphi:R\longrightarrow \bigoplus_{i=1}^n R/(q_i),$$ defined by $$\varphi(r)=(r+(q_1),...,r+(q_n))$$ where $q_i$ are coprime number. I have problem to show that it's surjective.
Let $(r_1+(q_1),...,r_n+(q_n))\in \bigoplus_{i=1}^n R/(q_i)$. How can I get $$(r_1+(q_1),...,r_n+(q_n))=(r+(q_1),...,r+(q_n))$$ for a certain $r\in R$ ?
Your map is surjective if and only if the ideals $(q_i)$ are pairwise coprime (Proposition 1.10 ii, in Atiyah and MacDonald). Try to prove it for $n=2$ first, that is: if $I,J$ are ideals of $R$ such that $I+J=R$, then $R \to R/I \oplus R/J$ is surjective. Here is a hint. Write $1=i+j, i\in I, j \in J$, and then, given $([x]_I,[y]_J)$, consider $$r \;:=\; xj+yi \;=\; x(1-i) + y(1-j).$$
When $n \geq 3$, fix $i$ between $1$ and $n$, and let $u_{i,j} \in (q_i), v_{i,j} \in (q_j)$ such that $u_{i,j} + v_{i,j} = 1$, when $i \neq j$. Then, given $([x_1]_{(q_1)}, \dots, [x_n]_{(q_n)})$, you can define
$$r \;:=\; x_1 a_1 + \dots + x_n a_n $$ with $1-a_i \in (q_i)$ such that $a_i \in \bigcap_{j \neq i} (q_j)$, for instance $$ a_i \;=\; \prod_{j \neq i} v_{i,j} \;=\; \prod_{j \neq i} (1-u_{i,j}) \;\equiv\; \prod_{j \neq i} 1 \equiv 1 \pmod{(q_i)} $$
(If $R$ is a PID, then any non-zero prime ideal is maximal, so $(q_i) + (q_j) = R$ as long as $i \neq j$ and the $(q_i)$'s are prime ideals. Therefore you can apply what I've said in the previous paragraph.)