Define $h: [0,1]\times [0,1]\rightarrow \mathbb{R}$, $h(x,y)= \cases{\frac{f(y)}{y} \,\, \text{if}\,\, y>x\\ 0 \,\, \text{otherwise}}$.
Then $\int\limits_{[0,1]} g \, d\lambda = \int\limits_{0}^{1} \int\limits_{x}^{1} \frac{f(y)}{y} \, dy \, dx = \int\limits_{0}^{1} \int\limits_{0}^{1} h(x,y) \, dy \, dx $, which by Fubini's theorem is
$\int\limits_{0}^{1} \int\limits_{0}^{1} h(x,y) \, dx \, dy = \int\limits_{0}^{1} \int\limits_{0}^{y} \frac{f(y)}{y} \, dx \, dy = \int\limits_{0}^{1} y\cdot \frac{f(y)}{y} \, dy = \int\limits_{0}^{1} f(y) \, dy = \int\limits_{[0,1]} f \, d\lambda$.
Note that $t \leq f(x) \leq 2t$ is equivalent to $f(x)/2 \leq t \leq f(x)$. In particular, we have $$f(x) = 2\int_{0}^{\infty} 1_{A_t}(x)~\mathrm dt. $$Using this, we obtain
\begin{align*}\int_{[0,1]} f(x) ~\mathrm dx &= \int_{\mathbb{R}^2}2\cdot\mathbb{1}_{[0,1]}(x)1_{A_{t}}(x)~ \mathrm dt ~\mathrm dx \\
&\leq\int_{\mathbb{R}^2}2\cdot g(x,t)~\mathrm dt ~\mathrm dx \\
&= \int_{[0,\infty)}\frac{2}{t}\left(\int_{A_{t}} f(x)~\mathrm dx\right)~\mathrm dt.\end{align*}
As $g(x,t) \geq 0$, exchanging the order of integration is easily justified using Fubini's theorem.
Best Answer
The key here is to notice that $$\int_{[0,1]}|f|\,d\lambda = \int_0^\infty \lambda(\{|f| > t\})\,dt = \sum_{n=1}^{\infty} \int_{2^{n-1}}^{2^{n}} \lambda(\{|f|>t\}) d t$$ Now notice that the function $t\mapsto \lambda(|f|>t)$ is decreasing and thus $$ 2^{n-1} \lambda(\{|f|>2^{n}\}) \leq \int_{2^{n-1}}^{2^{n}} \lambda(\{|f|>t\}) \,d t \leq 2^{n-1} \lambda(\{|f|>2^{n-1}\}) $$ for all $n\in \mathbb{Z}_{\geq 1}$. What you want to prove follows immediately from the above double inequality.