People usually write $\int f \mathrm d (\mu-\nu)$ as a shorthand for $\int f \mathrm d \mu -\int f \mathrm d \nu$.
Let $(\Omega, \mathcal F)$ be a measure space and $\mathcal M :=\mathcal M(\Omega)$ the space of all finite signed measures on $\Omega$. Let $\lambda \in \mathcal M$ and $(\lambda^+, \lambda^-)$ be its unique Jordan decomposition. Assume there are non-negative measures $\mu, \nu \in \mathcal M$ such that $\lambda = \mu-\nu$. Assume that $f$ is integrable w.r.t. both $\lambda^+$ and $\lambda^-$. Then we define $$ \int f \mathrm d \lambda := \int f \mathrm d \lambda^+- \int f \mathrm d \lambda^-. $$
Can we prove that $f$ is integrable w.r.t. both $\mu$ and $\nu$? If yes, can we show that $$ \int f \mathrm d \lambda = \int f \mathrm d \mu -\int f \mathrm d \nu $$ ?
Update: I have found a related question here.
No. $\lambda =(\mu+\tau)-(\nu +\tau)$ for any finite positive measure $\tau$ and we can certainly have $\int f d (\mu+\tau)=\infty$. I will let you write out an explicit counter-example.