So, I would like to prove "intrinsically" that if $g\in L^2(0,1)$ and $b > 0$, I can find $a\ge 0$ such that for all $f\in H^1(0,1) = W^{1,2}(0,1)$ $$ \|gf\|_2\,\le\,a\|f\|_2 + b\|f'\|_2. $$ I have a proof for this, but in the proof (see below) I extend everything to $\mathbb R$, prove it there and go back. By "intrinsically" I mean to stay on the interval. Does anybody have an idea or a reference?
My proof (sketch, not intrinsically):
(1) Proof of the analogous claim for $g\in L^2(\mathbb R)$ and $f\in H^1(\mathbb R)$. By Young's inequality, $\widehat{f'} = 2\pi i\omega\hat f$, and Cauchy-Schwarz, \begin{align*} \|gf\|_2 &= \|\hat g * \hat f\|_2\le\|g\|_2\|\hat f\|_1 = \|g\|_2\left(\int_{|\omega|<r}|\hat f|\,d\omega + \int_{|\omega|>r}\frac 1{|2\pi\omega|}|\widehat{f'}|\,d\omega\right)\\ &\le \|g\|_2\left(\sqrt{2r}\|f\|_2 + \frac{1}{\sqrt{2r}\pi}\|f'\|_2\right). \end{align*} (2) So, if $g\in L^2(0,1)$ and $f\in H^1_0(0,1)$, we can extend them by zero to $\mathbb R$ and get the desired property.
(3) Now, if $f\in H^1(0,1)$, let $h$ be the line connecting $(0,f(0))$ and $(1,f(1))$. Then $f-h\in H^1_0(0,1)$ and hence \begin{align*} \|gf\|_2 &\le\|g(f-h)\|_2 + \|gh\|_2\le a\|f-h\|_2 + b\|f'-h'\|_2 + \|g\|_2\|h\|_\infty\\ &\le (a\|f\|_2 + b\|f'\|_2) + a\|h\|_2 + b\|h'\|_2 + \|g\|_2\|h\|_\infty \end{align*} Each of $\|h\|_2$, $\|h'\|_2$, and $\|h\|_\infty$ is bounded by $|f(0)|+|f(1)|$. Now, we can use this and we are done.
Your linked question shows that for given $\epsilon > 0$ there exists $c \ge 0$ satisfying $$|f(x)| \le \frac{1}{4\epsilon} \|f\|_2 + \epsilon \|f'\|_2.$$ Thus $$|f(x) g(x)|^2 \le \left( \frac{1}{4\epsilon} \|f\|_2 + \epsilon \|f'\|_2 \right)^2 g(x)^2$$ so that $$\|fg\|_2 \le \left( \frac{1}{4\epsilon} \|f\|_2 + \epsilon \|f'\|_2 \right)\|g\|_2.$$
If $\|g\|_2 = 0$ there is nothing to show; otherwise select $\epsilon = \dfrac{b}{\|g\|_2}$.