I'm unsure how I arrived at this assumption but I assumed :
$$ \frac{a}{a+b} = \frac{a}{a} + \frac{a}{b} $$
testing with values $a = 3$ and $b = 4$ this is not true as
$$ \frac{3}{3+4} \neq \frac{3}{3} + \frac{3}{4} $$
Can $\frac{a}{a+b}$ be expanded so that $a$ exclusively is contained in its own term and $b$ is contained in its own term.
Your assumption, as you note, is not true at all. In particular, if you add those two fractions together and simplify a little, you get $\frac{a + b}{b}$. No such decomposition is possible.
In particular, you can never hope to express $\frac{a}{a+b}$ (or any multiple thereof as a sum of something with an $a$ in the denominator and something with a $b$ in the denominator, because $a+b$ is not a factor of $ab$ in general, but $\frac{c}{a} +\frac{d}{b} = \frac{bc+ad}{ab}$, so the denominator of it must be some factor of $ab$.