Too long for a comment: Your formula $$\frac{1}{c^2}L(cB)=\frac{1}{c^2}\left(F(cB(T))-\frac{c^2}{2}\int_0^T f'(cB(t))dt-\frac12\int_0^T f^2(cB(t))dt\right):=\frac{1}{c^2}L^c(cB).$$ I cannot reproduce.

When I integrate $\int_0^Tf(cB(t))\,d(cB(t))$ by parts I get \begin{align} &\int_0^Tf(cB(t))\,d(cB(t))=-\int_0^TcB(t)\,df(cB(t))+\Big\langle f(cB),cB\Big\rangle_T\\ &\stackrel{\text{Ito's f.}}{=}-\int_0^TcB(t)\,f'(cB(t))\,d(cB(t))-\frac{1}{2}\int_0^TcB(t)\,f''(cB(t))\,c^2\,dt+\Big\langle f(cB),cB\Big\rangle_T\,. \end{align} Using Ito's formula we also get $$ \Big\langle f(cB),cB\Big\rangle_T=\int_0^Tf'(cB(t))\,dt\,. $$ Putting this together it looks quite different from your $L(cB(t)).$

In a nutshell you are asking the following:

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 $c^2$.