Assume $\left|G \right|= \left| G \right| \cdot 1_F$ is invertible in $F$. Let $\Delta:F \to GL_n(F)$ be a representation and $U \subseteq F^n$ be an $F$-subspace that is $\Delta$-invariant. Then there exists a $\Delta$-invariant $F$-subspace $V \subseteq F^n$ such that $F^n= U \oplus V$.
Proof. Let $u_1, \ldots, u_r$ be an $F$-basis of $U$ and extend it to an $F$-basis $u_1,\ldots,u_n$ of $F^n$. Let $A \in \text{Mat}_n(F)$ be defined by $Au_i = u_i$ for $i=1,\ldots,r$ and $Au_i=0$ for $i=r+1,\ldots,n$.
We set $A^{\prime} := \frac{1}{\left|G \right|} \displaystyle \sum_{g \in G} \Delta(g^{-1}) A \Delta(g)$.
Question $1$: What is the motivation for introducing $A^{\prime}$? It seems to pop up out of nowhere (Its use in the proof in not confusing, just not sure where it comes from)
Then we have: (i). $A^{\prime} u = u$ for every $u \in U$.
(ii). $A^{\prime} x \in U$ for every $x \in F^n$.
(iii). $A^{\prime} A^{\prime} x = A^{\prime} x$ for every $x \in F^n$
(iv). $A^{\prime} \Delta(h) = \Delta(h) A^{\prime}$ for every $g \in G$.
Question $2$: I understand how the text proves all except (ii). It states, to see (ii), note that $A \Delta(g) x \in U$. I can understand that $A \Delta (g) \in \text{Mat}_n(F)$ and $x \in F^n$, so $A \Delta (g) x \in F^n$, but why in $U \subseteq F^n$? (Edit: No longer need explanation, answered in comments)
Q1: The point is that $G$-homomorphisms $M \to M$ are exactly the same thing as elements of $\hom_F(M,M)$ on which $G$ acts trivially. Notice that in $FG$, for any $x\in G$, $$x \sum_{g \in G}g = \sum_{g \in G}g.$$ So given any $FG$-module $N$ and $n \in N$, the element $\left(\sum_{g \in G} g\right)n= \sum_{g \in G} gn$ is either zero or spans a trivial submodule. That weird looking thing is what you get by applying $\sum g$ to the element $A$ of the $FG$-module $\hom_F(F^n,F^n)$ (think about the definition of the module action on $\hom_F$), so it is a $G$-hom.