Failure of the Krull-Schmidt Theorem? Theorem 1.19 of  Representation Theory of Finite Groups: Algebra and Arithmetic is the Krull-Schmidt theorem, which I screenshotted and uploaded it here, I don't have any problem with this theorem and the proof is clear to me.  

But I have some questions about the following remark, which follows the theorem,

I know why $R$ is free, but I wonder why $M$ is not free? some information is given but I am not convinced and I cannot see the details.
Again I know why $R$ is indecomposable as a left $R$- module, but I wonder why $M$ is also indecomposable as left $R$- module? 
And finally the final isomorphism is not clear to me, can anyone please give me some details? 
Thanks in advance for any help! 
 A: The hint they give isn't aimed at showing that $M$ isn't free, rather that it isn't free of rank one.  The point is that if $m \in M$ then the submodule $Rm$ cannot be all of $M$: there is some $x \in [0,1]$ with $m(x)=0$ by the intermediate value theorem, thus every element of $Rm$ vanishes at $x$, and it follows $Rm \neq M$ as $M$ contains functions not vanishing at $x$.
It is true that $M$ is not free.  We've already shown it is not free of rank one, and we show it is not free of any higher rank by proving any two elements are not $R$-linearly independent.  Let $m,n \in M$ be nonzero, let $0\neq g \in R$ such that $g(0)=g(1)=0$ and $gm,gn \neq 0$, and note $gm,gn \in R$.  Then $(gm)n = (gn)m$ which shows $m,n$ are not $R$-linearly independent.
I think a similar argument will show that $M$ is indecomposable by showing any two nonzero submodules have a nonzero element in common.
To understand the definition of the isomorphism $\varphi$, identify $R\oplus R$ with the continuous functions $a: [0,1]\to \mathbb{R}^2$ such that $a(0)=a(1)$.  The pair $(r,s) \in R\oplus R$ corresponds to the function $x \mapsto (r(x),s(x))$ under this identification.  Then if $f,g \in M$, $\varphi(f,g)$ is the function $[0,1] \to \mathbb{R}^2$ sending $x$ to the point of $\mathbb{R}^2$ obtained by rotating $(f(x),g(x))$ by $\pi x$ about the origin.  This is continuous as a composition of continuous functions, and it satisfies the endpoint condition: $0$ gets sent to $(f(0),g(0))$, and $1$ to $(f(1),g(1))= (-f(0),-g(0))$ rotated by $\pi$, which is $(f(0),g(0))$.
