Consider the sum $f(\frac{a}{a + b}) + f(\frac{b}{a + b})$ with a function $f: \Bbb R \to \Bbb R$ and $a, b \in \Bbb R$. I am trying to find all functions $f$ for which one can find a function $g: \Bbb R \to \Bbb R$ such that the functional equation $f(\frac{a}{a + b}) + f(\frac{b}{a + b}) = g(a + b)$ is fulfilled. I would expect that one can only find such a function $g$ for linear functions $f$.
That one can find a function $g$ for linear functions $f$ can be easily proven as follows. Consider $f(x) = \alpha (x - x_0) + y_0$ with $\alpha, x_0, y_0 \in \Bbb R$. It holds \begin{align} f \Big (\frac{a}{a + b} \Big ) + f \Big (\frac{b}{a + b} \Big ) & = \alpha \Big (\frac{a}{a + b} - x_0 \Big ) + y_0 + \alpha \Big (\frac{b}{a + b} \Big ) + y_0 \\ & = \alpha (1 - 2 x_0) + 2 y_0 \\ & = g(a + b) \end{align}
Can it be proven that one can only find such a function $g$ for linear functions $f$?
Every function such that $f(x)+f(1-x)=g$ (where $g$ is a constant) satisfies your conditions. There are plenty such functions, namely all those whose graph passes through $(\frac12, \frac12g)$ and is symmetric with respect to that point, for instance $f(x)=(x-\frac12)^3+\frac12g$.
(On the other hand taking $a=b$ you see that the function $g$ must be constant, at least except from $0$, so the first paragraph should characterize all solutions.)