By definition, $\operatorname{Rad}(M)$ equals the intersection of all submodules $N$ of $M$ such that $M/N$ is simple.
Let $N$ be a submodule of $M$ such that $M/N$ is simple. Then there exists $\mathfrak m\in\operatorname{Max}(R)$ such that $M/N\simeq R/\mathfrak m$. In particular, $\mathfrak m(M/N)=0$ and thus $\mathfrak mM\subseteq N$. This shows that $\bigcap_{\mathfrak m\in \Omega}\mathfrak mM\subseteq \operatorname{Rad}(M)$.
For the converse note that $\operatorname{Rad}(M)$ also equals the intersection of all submodules $N$ of $M$ such that $M/N$ is semisimple. Since $M/\mathfrak mM$ is an $R/\mathfrak m$-vectorspace, then $M/\mathfrak mM$ is semisimple and we are done.
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
Note that the ring $R$, and each of the rings $R/I_k$, is also an $R$-module. It is easy to check that the map $\varphi$ is an $R$-module homomorphism.
By the definition of "comaximal", any given maximal ideal $M\subset R$ will contain at most one $I_k$.
For any given maximal ideal $M\subset R$ and any $I_k$, we have $(R/I_k)_M\cong R_M/(I_k)_M$ because localization commutes with taking quotients. If $M$ contains $I_k$, then $(I_k)_M\neq R_M$, whereas if $M$ does not contain $I_k$, we have $(I_k)_M=R_M$ because $I_k\cap (R\setminus M)\neq\varnothing$.
In either case, we see that the localized map $$\small\varphi_M:R_M\to \left(\prod R/I_k\right)_M\cong \prod R_M/(I_k)_M\cong \begin{cases} R_M/(I_k)_M &\text{if $I_k$ is contained in $M$,}\\ 0 & \text{if none of the $I_k$'s are contained in $M$} \end{cases}$$ is just a quotient map from the ring $R_M$, hence surjective.
(I leave it to you to check that the composition of $\varphi_M$ with all of these isomorphisms actually does result in the quotient map $R_M\to R_M/(I_k)_M$.)
P.S. This is actually the first time I've seen this approach to proving the map in the Chinese remainder theorem is surjective. I like it a lot, though there are proofs which don't need any localization. Consider the approach taken in Atiyah-Macdonald: