Consider, for $m\ge2$, the function $\varphi_m(x):=x(1-x)^m$ . So 
$$\varphi_m(0)=0\qquad \varphi_m(1)=0$$  $$ \varphi_m'(0)=1\qquad  \varphi_m'(1)=0\ .$$ It is also easy to see that  $\|  \varphi_m'\|_2=O(m^{-{1/2}})$ (in fact, we may compute exactly both $\|  \varphi_m\|_2^2$ and $\| \varphi_m'\|_2^2$ in terms of some rational functions  of $m$, by means of a few Beta function integrals).

For any $v\in H^2(0,1)$ define

$$v_m(x):=v(x)+\big(v(0)-v'(0)\big)\varphi_m(x)+\big(v(1)+v'(1)\big)\varphi_m(1-x)$$

so $$v_m(0)=v(0)\qquad v_m(1)=v(1)$$ $$v_m'(0)=v(0)\qquad v_m'(1)=-v(1)$$

that is $v_m\in V$; clearly $\|v-v_m\|_{1,2}=O(m^{-{1/2}})$.