Let $f'\in\mathcal{R}[0,1]$ then prove the above inequality.
Try : What I have found till now is that $$\left|\frac{f(1)+f(0)}{2}\right|\leq \int_{0}^{1}|f(t)|dt+\frac{1}{2}\int_{0}^{1}|f'(t)|dt$$ and $$\left|\frac{f(1)}{2}\right|\leq \int_{0}^{1}\left|f(t)\right|dt+\frac{1}{2}\int_{0}^{1}|f'(t)|dt.$$ By using the integral $\int tf(t)dt$ and $\int (1-t)f(t)dt$. But I don't understand is that how do I get the $f(\frac{1}{2})$?
Your idea is essentially right, we just need to use different boundaries and integration by parts: $$\int_0^{1/2}tf'(t)dt=[tf(t)]^{1/2}_0-\int_0^{1/2}f(t)dt$$ so $$\frac{f(1/2)}{2}=\int_0^{1/2}f(t)dt+\int_0^{1/2}tf'(t)dt.\tag{1}$$ Now do the same analysis with $g(t)=f(1-t)$: $$\frac{f(1/2)}{2}=\frac{g(1/2)}{2}=\int_0^{1/2}g(t)dt+\int_0^{1/2}tg'(t)dt=\int_{1/2}^1f(t)dt+\int_{1/2}^1(1-t)f'(t)dt\tag{2}$$ Add (1) and (2): \begin{align*} |f(1/2)|&=|\int_0^1 f(t)dt+\int_0^{1/2}tf'(t)dt+\int_{1/2}^1(1-t)f'(t)dt|\\ &\leq\int_0^1|f(t)|dt+\int_0^{1/2}|t||f'(t)|dt+\int_{1/2}^1|1-t||f'(t)|dt\\ &\leq\int_0^1|f(t)|dt+\frac{1}{2}\int_0^1|f'(t)|dt \end{align*}