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/ .
Best Answer
This came up before on MO, and the sentiment was that Rota is referring to the Elementwise Chinese Remainder Theorem. This holds in a ring $R$ if for any ideals $I_1,\ldots,I_n$ of $R$ and elements $x_1,\ldots,x_n \in R$, the following are equivalent:
(i) $x_i - x_j \in I_i + I_j$.
(ii) There is $x \in R$ with $x \equiv x_i \pmod{I_i}$ for all $i$.
It is clear that (ii) $\implies$ (i) in any ring. It turns out that the domains in which (i) $\implies$ (ii) holds are precisely the Prüfer domains. You can read a little bit about Prüfer domains in $\S 21$ of my commutative algebra notes, but not as much as I would like: this is the point at which the notes begin to trail off. In particular, the above result appears in the notes but the proof does not! (I think it appears in Larsen and McCarthy's Multiplicative Ideal Theory, for instance.)
Added much later: The proof does appear there now.