It seems that the OP is confusing signed measure by Radon measure.
In general, if $f\in L^{loc}_1(\mathbb{R})$, then $\nu_f(dx)=f(x)\,dx$$\nu^f(dx)=f(x)\,dx$ is not a signed-measure for $\mu(\mathbb{R})$ may be undefined. Example: $f(x)=x$; $g(x)=\sin x$, etc.
When consider as a functional on $C_{00}(\mathbb{R})$ however, $\nu_f$$\nu^f$ satisfies the following property:
Property R: For any sequence $\{\phi_n:n\in\mathbb{N}\}\subset\mathbb{C}_{00}(\mathbb{R})$ that is supported an a compact set $K$ (i.e., $\operatorname{supp}(\phi_n)\subset K$ for all $n\in\mathbb{N}$) if $\phi_n\xrightarrow{n\rightarrow\infty}\phi$ uniformly on $\mathbb{R}$, then $$\begin{align}\nu_f(\phi_n)\xrightarrow{n\rightarrow\infty}\nu_f(\phi)\tag{R}\label{R}\end{align}$$$$\begin{align}\nu^f(\phi_n)\xrightarrow{n\rightarrow\infty}\nu^f(\phi)\tag{R}\label{R}\end{align}$$
Functionals $\nu$ that satisfy \eqref{R} are called Radon measures A decomposition similar to the Hahn decomposition for signed measures exists for Radon measures or general $\sigma$-continous elementary integrals with finite variation (see Bichteler, K., Integration: A functional approach. Birkhäuser Advanced Texts Basler Lehrbücher. 1998th Edition).
Observation: A regular signed measure is a Radon measure; a Radon measure is not necessarily a signed measure. The restriction of a Radon measure to a compact set $K$ defines a (signed) measure supported in $K$.
Now, if $F$ and $G$ are functions on $\mathbb{R}$ of local bounded variation (i.e. $F$, and $G$ have finite variation on any bounded closed interval $[a,b]$ then the Lebesgue integration by parts formula gives $$\begin{align} \int_{(a,b]}F(t)\mu_G(dt)=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t-)\mu_F(dt)\tag{1}\label{one}\end{align}$$ where $\mu_F$ and $\mu_G$ are the Radon measures induced by $F$ and $G$ (i.e. $\mu_F((a,b])=F(b)-F(a)$, and $\mu_G((a,b])=G(b)-G(a)$ for all $-\infty<a\leq b<\infty$). If $F\in\mathcal{C}^\infty_{00}(\mathbb{R})$$\phi=F\in\mathcal{C}^\infty_{00}(\mathbb{R})$, then $\mu_F(dx)=F'(x)\,dx$$\mu_\phi(dx)=\phi'(x)\,dx$ and \eqref{one} takes the form $$\begin{align} \int_{(a,b]}F(t)\mu_G(dt)=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t-)F'(t)\,dt\tag{2}\label{two} \end{align}$$$$\begin{align} \int_{(a,b]}\phi(t)\mu_G(dt)=\phi(b)G(b)-\phi(a)G(a)-\int_{(a,b]}G(t-)\phi'(t)\,dt\tag{2}\label{two} \end{align}$$ If $\operatorname{supp}(F)\subset [a,b]$$\operatorname{supp}(\phi)\subset [a,b]$, then $\mu_F(dx)=F'(x)\,dx$$\mu_\phi(dx)=\phi'(x)\,dx$ and \eqref{two} becomes $$\begin{align} \int_\mathbb{R} F(t)\,\mu_G(dt)=\int_{(a,b]}F(t)\mu_G(dt)=-\int_{(a,b]}G(t-)F'(t)\,dt=-\int_{\mathbb{R}}G(t-)F'(t)\,dt\tag{3}\label{three} \end{align}$$$$\begin{align} \int_\mathbb{R} \phi(t)\,\mu_G(dt)&=\int_{(a,b]}\phi(t)\mu_G(dt)\\ &=-\int_{(a,b]}G(t-)\phi'(t)\,dt=-\int_{\mathbb{R}}G(t-)\phi'(t)\,dt\tag{3}\label{three} \end{align}$$ If $G(t)=\int^t_0 g(s)\,dx$ where $g\in L^{loc}_1(\mathbb{R})$, meaning $G(t)=\int_{[0,t]} g(s)\,dx$ if $t\geq0$ and $-\int_{[t,0]}g(s)\,ds$ when $t<0$, then \eqref{three} takes the form $$\begin{align} \int_\mathbb{R} F(t)\,g(t)\,dt=-\int_{\mathbb{R}}G(t)F'(t)\,dt\tag{4}\label{four} \end{align}$$$$\begin{align} \int_{\mathbb{R}}\phi(t)\,\mu_G(dt)=\int_\mathbb{R} \phi(t)\,g(t)\,dt=-\int_{\mathbb{R}}G(t)\phi'(t)\,dt\tag{4}\label{four} \end{align}$$