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 $c^2$.