$H \subseteq \textrm{ker } \Psi$ is all you need for $\Psi$ to factor through the quotient $F/H$.
In general, let $\delta: A \rightarrow B$ be a homomorphism of abelian groups ($\mathbf Z$-modules, same thing), and let $A_0$ be a subgroup of $A$. To say that $\delta$ factors through $A_0$ (or $A/A_0$) is to say that there exists an abelian group homomorphism $\bar{\delta}: A/A_0 \rightarrow B$ such that $\bar{\delta}(a + A_0) = \delta(a)$ for all $a \in A$.
If such a homomorphism exists, it is clearly unique. The only issue barring the existence of such a homomorphism is the possibility that it is not well defined. And you can check that it is well defined if and only if $A_0$ is contained in the kernel of $\delta$.
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
For element $x\in S+T$, you can write $x=s+t$ for $s\in S$ and $t\in T$, and then define $\psi(x)=s+S\cap T$. We have to check that this does not depends on the choice of $s$ and $t$, so if $x=s'+t'$ for $s'\in S$ and $t'\in T$, from $x=s+t=s'+t'$ we have $s-s'=t'-t$ which belongs to $S\cap T$. Hence $s+ S\cap T= s'+S\cap T$, and $\psi(x)$ does not depend on the choice of $s$ and $t$.
This is clearly a homomorphism. You can easily check that $\ker(\psi)=T$.