Let $f$ be of bounded variation on $[a,b]$. On this interval, it can be shown that $f$ can be written as a difference between two increasing functions. It can be shown that these two function can be chosen so that : $f^+=f(x)+TV(f_{[a,x]})$, and $f^-=TV(f_{[a,x]})$. We then have that $f=f^+-f^-$.
According to this link: link.
$TV(f_{[a,b]})=f^+(b)-f^+(a)+f^-(b)-f^-(a)$.
But if I use the representation above I get that:
$f^+(b)-f^+(a)+f^-(b)-f^-(a)=f(b)+TV(f_{[a,b]})- \\f(a)-TV(f_{[a,a]})+TV(f_{[a,b]})-TV(f_{[a,a]})=2TV(f_{[a,b]})+f(b)-f(a)$
And it may not be that $2TV(f_{[a,b]})+f(b)-f(a)=TV(f_{[a,b]})$?
Do you see where my mistake is?
You misunderstood what the link says. The link says that there is a pair $(f^+,f^-)$ of increasing functions with $f = f^+ - f^-$ such that the equality
$$TV(f_{[a,b]}) = f^+(b) - f^+(a) + f^-(b) - f^-(a)\tag{1}$$
holds (and that this pair is unique up to addition of constants). That doesn't work for all pairs $(g,h)$ of increasing functions such that $f = g-h$, and the particular decomposition you have isn't mentioned in that section.
The standard choice for $f^+,f^-$ satisfying $(1)$ is
$$f^+(x) = \frac{TV(f_{[a,x]}) + f(x)}{2}\quad\text{and}\quad f^-(x) = \frac{TV(f_{[a,x]}) - f(x)}{2},$$
for which $(1)$ is easily verified. Condition $(1)$ is a minimality condition, increasing functions $g,h$ with $f = g-h$ have the minimal possible variation when $(1)$ holds. Generally, one has
$$TV(f_{[a,b]}) \leqslant g(b) - g(a) + h(b) - h(a)$$
for all increasing $g,h$ with $f = g-h$.
The choice $\bigl(f(x) + TV(f_{[a,x]}), TV(f_{[a,x]})\bigr)$ only satisfies $(1)$ if $f$ is nonincreasing.