Let $(\Omega,B,\mu)$ be a probability space where $\Omega$ is $[0,1]$, $\mu$ the Lebesgue measure, $B$ the Borel $\sigma$-algebra of $[0,1]$ and $f(w)=1-w$ be a random variable. Let $\phi:\mathbb R\rightarrow\mathbb R$ and $\phi(x) = X$ Our definition of expectation is:
$$E(f) = \int_\Omega f(w) \mu(dw),$$
and by the measure transformation theorem: $$E(f)=\int_R \phi(f(w))\mu(dw) = \int_Rx\mu_f(dx).$$
Using this definition and:
$$\mu_f=\begin{cases} \mu(\emptyset) = 0 & x < 0 \\ \mu([1-x,1)) = x & 0 \leq x \leq 1\\ \mu([0,1]) = 1 & x > 1 \end{cases}.$$
Is the expectation of $f(w)=1-w$ just:
$$E(f) = \int_{\mathbb R}x\mu_f(dx) = \int_{[0,\infty)}x\mu_f(dx) - \int_{(\infty,0]}x\mu_f(dx),$$
where:
$$\int_{[0,\infty)}x\mu_f(dx) = x\mu_f([0,1]) + x\mu_f((1,\infty)) = x*x + x?$$
The end of your calculation is wrong, the expectation is a number, not a function of $x$.
Here you don't need to introduce the law of $f$ because you know $(\Omega,B,\mu)$. So if you want to compute the expectation you just need to use the definition that you recalled.
$$E(f)=\int_\Omega fd\mu=\int_0^1(1-\omega)d\omega=\frac{1}{2}$$
Now if you really want to use $\mu_f$, your calculation of the cdf is correct: $$\mu_f((-\infty,x])=\mu(f\leq x)=\left\lbrace\begin{array}{cc} 0, & x<0\\ x, & 0\leq x\leq 1\\ 1, & 1<x \end{array}\right.$$ So $\mu_f$ has a density w.r.t. the Lebesgue measure, which is $\mathbf{1}_{[0,1]}$, i.e. $\mu_f(dx)=\mathbf{1}_{[0,1]}(x)dx$. Then the expectation is $$E(f)=\int_\mathbb{R}x\mu_f(dx)=\int_\mathbb{R}x\mathbf{1}_{[0,1]}(x)dx=\int_0^1xdx=\frac{1}{2}$$