Where does the extra term in the density of a diffusion with respect to $c B(t)$ come from?

It is well known that for the diffusions

\begin{align*} dX&=f(X)dt+&cdB\\ dY&=&cdB \end{align*}

the density of the law of $$X$$ with respect to the law of $$Y$$ is

\begin{align*} \frac{d\mu_X}{d\mu_Y}(c B)&=\exp\left(\int_0^T\frac{f(cB(t))}{c}dB(t)-\frac12\int_0^T\frac{f^2(cB(t))}{c^2}dt\right)\\ &=\exp\left(\frac{1}{c^2}\left(\int_0^Tf(cB(t))d(cB(t))-\frac12\int_0^Tf^2(cB(t))dt\right)\right)\\ &:=\exp\left(\frac{1}{c^2}L(cB(t))\right),\end{align*}

where $$L(cB)=\int_0^Tf(cB(t))d(cB(t))-\frac12\int_0^Tf^2(cB(t))dt$$. My interest now is in making $$\frac{1}{c^2}L(cB(t))$$ defined pathwise. Applying integration by parts yields that

$$\frac{1}{c^2}L(cB)=\frac{1}{c^2}\left(F(cB(T))-\frac{c^2}{2}\int_0^T f'(cB(t))dt-\frac12\int_0^T f^2(cB(t))dt\right):=\frac{1}{c^2}L^c(cB).$$

I guess my question is - why does the "pathwise" functional depend on $$c$$ where the nonpathwise functional doesn't? What is the term $$\int_0^T f'(B(t))dt$$ in relation to the original $$L$$? Is there a nice expression?

If we define $$L_1(B)=\int_0^Tf(B(t))d(B(t))-\frac12\int_0^Tf^2(B(t))dt$$ and $$L_2(B)= F(B(T))-\frac{1}{2}\int_0^T f'(B(t))dt-\frac12\int_0^T f^2(B(t))dt\$$. Then $$L_1(B)=L_2(B)$$ however seemingly $$L_1(cB)\neq L_2(cB)$$. Edit:actually this comes from the fact that Ito's lemma only holds a.s. and cB is singular with B for $$c\neq \pm 1$$.

When $$F(x)=\int_0^xf(y)\,dy$$ we get from Ito $$F(cB(t))=\int_0^Tf(cB(t))\,d(cB(t))+\frac{1}{2}\int_0^Tf'(cB(t))\,c^2\,dt\,$$ so that the expressions \begin{align} &F(cB(t))-\frac{c^2}{2}\int_0^Tf'(cB(t))\,dt\,,\tag{1}\\ &\int_0^Tf(cB(t))\,d(cB(t))\tag{2} \end{align} are equal. Then why (2) depends only on $$cB(t)$$ and not separately on $$c^2$$ and $$cB(t)$$ like (1) seems to do?
Good question. I think the answer is that (1) can also be written as \begin{align} &F(cB(t))-\frac{1}{2}\int_0^Tf'(cB(t))\,d\big\langle cB\big\rangle_t\,,\tag{1'} \end{align} and gone is the isolated $$c^2$$.