In the following I am assuming that $\mathbb T=S^1$ is just the circle, but I am just unfamiliar with this notation. My task is as follows
Let $f\in L^2(\mathbb T)$ be differentiable when restricted to the interval $(-1/2,1/2)$ with bounded derivative.
Show that $f=g-h$ where both $g$ and $h$ are non-decreasing differentiable functions in $(-1/2,1/2)$.
I am thinking that the assumption $f\in L^2(\mathbb T)$ reduces to $f\in L^2([-1/2,1/2])$ with $f(-1/2)=f(1/2)$, doesn't it?
So here is my attempt:
Wlog assume $f(-1/2)=f(1/2)=0$.
Let $f'_+(x)=\max(f'(x),0)$ and $f'_-(x)=f'_+(x)-f'(x)$.
Note that $f'_-(x)\geq 0$.
Now define
$g(x)=\int_{-1/2}^x f'_+(x)dx$ and
$h(x)=\int_{-1/2}^x f'_-(x)dx$.
Then
$$g(x)-h(x)=\int_{-1/2}^x f'(x)dx=f(x)$$
There are several points that need clarification. There is no need to assume that $f$ vanishes at $\pm \frac 1 2$. (You have to justify 'WLOG ...). Secondly, the Fundamental Theorem of Calculus is usually proved under the assumption that $f'$ is continuous. Though it is applicable here you need a good knowledge of measure theory to justify it. So it is better to use the fact that any function of bounded variation can be written as the difference of two monotone functions. The fact that our $f$ is of bounded variation follows easily from MVT and the fact that $f'$ is bounded.