Quasi-Chinese Remainder Theorem for modules Let $R$ be a ring. The Chinese Remainder Theorem asserts that if $J_1,\ldots,J_n\lhd R$ are bilateral coprime ideals (i.e. $J_i+J_k=R,\ \forall i\neq k$) then the map $$\varphi:a\in R\longmapsto (a+J_1,\dots,a+J_n)\in R/J_1\oplus\dots\oplus R/J_n $$ is a surjective ring morphism, such that $$\ker (\varphi)=\displaystyle \bigcap _{1\leq i\leq n} J_i.$$  Thus we have a ring isomorphism (and thus an $R$-module isomorphism) $$R/\bigcap_{1\leq i\leq n} J_i\simeq R/J_1\oplus\dots\oplus R/J_n.$$ 
In the proof the fact that the ideals are bilateral is used not only to make sense of the ring quotient $R/J_i$. 
So I’m wondering whether I could extend the theorem in order to obtain a module isomorphism instead of a ring one in the following sense:

Let $R$ be a non-commutative ring (the commutative case is in fact the Chinese Remainder Theorem itself) and $J_1,\dots,J_n\subset R$ be left ideals such that $J_i+J_k=R$ if $i\neq k$. Let $R/J_i$ be the quotient module of $R$ modulo the left $R$-submodule $J_i$. Then the mapping $\varphi$ (defined above) is a surjective $R$-module morphism. 

The problem arose in trying to prove (or disprove) that any finite dimensional $k$-algebra has only a finite number of simple left modules up to isomorphism. If the module-generalisation theorem holds, then I have proved this fact. 
 A: The answer that follows is just a detailed version of my comments to the
original post.
Your conjecture (that $\varphi$ is surjective) is true for each $n\leq2$ but
false for each $n\geq3$. Let me show this. First, let me prove that it is true
for $n=2$ (for the sake of completeness -- I know that you have a proof):

Proposition 1. Let $R$ be a (noncommutative, associative, unital) ring.
  Let $U$ and $V$ be two left ideals of $R$ such that $U+V=R$. Let
  $\varphi:R\rightarrow\left(  R/U\right)  \oplus\left(  R/V\right)  $ be the
  map that sends each $r\in R$ to $\left(  r+U,r+V\right)  \in\left(
R/U\right)  \oplus\left(  R/V\right)  $. Then, $\varphi$ is a surjective
  $R$-module homomorphism.

Proof of Proposition 1. It is clear that $\varphi$ is an $R$-module
homomorphism. It thus remains to prove that $\varphi$ is surjective.
We have $1\in R=U+V$. In other words, there exist $u\in U$ and $v\in V$ such
that $1=u+v$. Consider these $u$ and $v$.
Let $z\in\left(  R/U\right)  \oplus\left(  R/V\right)  $ be arbitrary. Thus,
we can write $z$ in the form $z=\left(  \alpha,\beta\right)  $ for some
$\alpha\in R/U$ and $\beta\in R/V$. Consider these $\alpha$ and $\beta$.
We have $\alpha\in R/U$; thus, there exists some $a\in R$ such that
$\alpha=a+U$. Consider this $a$.
We have $\beta\in R/V$; thus, there exists some $b\in R$ such that $\beta
=b+V$. Consider this $b$.
Let $x=av+bu\in R$. Thus,
\begin{align*}
\underbrace{x}_{=av+bu}-\underbrace{a}_{\substack{=a1=a\left(  u+v\right)
\\\text{(since }1=u+v\text{)}}}  & =av+bu-a\left(  u+v\right)  =bu-au=\left(
b-a\right)  \underbrace{u}_{\in U}\\
& \in\left(  b-a\right)  U\subseteq U\qquad\left(  \text{since }U\text{ is a
left ideal of }R\right)  ,
\end{align*}
so that $x+U=a+U=\alpha$. Also,
\begin{align*}
\underbrace{x}_{=av+bu}-\underbrace{b}_{\substack{=b1=b\left(  u+v\right)
\\\text{(since }1=u+v\text{)}}}  & =av+bu-b\left(  u+v\right)  =av-bv=\left(
a-b\right)  \underbrace{v}_{\in V}\\
& \in\left(  a-b\right)  V\subseteq V\qquad\left(  \text{since }V\text{ is a
left ideal of }R\right)  ,
\end{align*}
so that $x+V=b+V=\beta$. Now, the definition of $\varphi$ yields
\begin{equation}
\varphi\left(  x\right)  =\left(  \underbrace{x+U}_{=\alpha},\underbrace{x+V}
_{=\beta}\right)  =\left(  \alpha,\beta\right)  =z.
\end{equation}
Thus, $z=\varphi\left(  \underbrace{x}_{\in R}\right)  \in\varphi\left(
R\right)  $.
Now, forget that we fixed $z$. Thus, we have shown that $z\in\varphi\left(
R\right)  $ for each $z\in\left(  R/U\right)  \oplus\left(  R/V\right)  $. In
other words, the map $\varphi$ is surjective. This completes the proof of
Proposition 1. $\blacksquare$
Proposition 1 shows that your conjecture holds for $n=2$. For $n<2$, it is
completely obvious (since $\varphi$ is a projection map in this case). Now,
let me disprove your conjecture for $n>2$ using the following example:

Example 2. Let $n>2$ be an integer. Let $V$ be the $2$-dimensional
  $\mathbb{Q}$-vector space $\mathbb{Q}^{2}$. For each positive integer $i$, let
  $e_{i}$ be the vector $\left(  1,i\right)  ^{T}\in V$. Note that these vectors
  $e_{1},e_{2},e_{3},\ldots$ are pairwise linearly independent. Let
  $R=\operatorname*{End}\nolimits_{\mathbb{Q}}V\cong\mathbb{Q}^{2\times2}$. For
  each positive integer $i$, let $J_{i}$ be the subset $\left\{  A\in
R\ \mid\ Ae_{i}=0\right\}  $ of $R$. It is clear that all these subsets
  $J_{1},J_{2},J_{3},\ldots$ are left ideals of $R$. Moreover, each $J_{i}$ is a
  $2$-dimensional subspace of the $4$-dimensional $\mathbb{Q}$-vector space $R$.
  But any two positive integers $i$ and $j$ satisfy $J_{i}\cap J_{j}=0$ (because
  any endomorphism $A\in R$ that annihilates the two linearly independent
  vectors $e_{i}$ and $e_{j}$ must be the zero map) and therefore $J_{i}
+J_{j}=R$ (since $\dim\left(  J_{i}+J_{j}\right)  =\underbrace{\dim\left(
J_{i}\right)  }_{=2}+\underbrace{\dim\left(  J_{j}\right)  }_{=2}
-\underbrace{\dim\left(  J_{i}\cap J_{j}\right)  }_{=0}=4$). Hence, if your
  conjecture were true, the map $\varphi:R\rightarrow\left(  R/J_{1}\right)
\oplus\left(  R/J_{2}\right)  \oplus\cdots\oplus\left(  R/J_{n}\right)  $
  would be surjective. This would yield $4\geq2n$, since this map $\varphi$ is a
  $\mathbb{Q}$-linear map from a $4$-dimensional $\mathbb{Q}$-vector space to a
  $\left(  2n\right)  $-dimensional $\mathbb{Q}$-vector space; but this would
  contradict $n>2$.

Finally, the question you were trying to solve in the last paragraph is a
consequence of Theorem 1 in https://mathoverflow.net/a/14516/ .
