Problem: 
My Attempt: Let $P_{BD}$ represent the change of coordinates from the basis $B$ to $D.$ Then we have that $$(v)_{B}=P_{BD}(v)_{D}.$$ Assuming $$P= \left( \begin{array}{ccc} x_1 & y_1 & z_1 & w_1\\ x_2 & y_2 & z_2 & w_2\\ x_3 & y_3 & z_3 & w_3\\ x_4 & y_4 & z_4 &w_4 \end{array} \right)$$ then on solving for $P$ we get that $x_1+2z_1=0 $, $x_2+2z_2=1$, $x_3+2z_3=1$ and $x_4+2z_4=2.$ Choosing $z_1=1,z_2=0,z_3=0$ and $z_4=0$ we get that $x_1=-2,x_2=1,x_3=1,x_4=1.$ I guess that we can make choices for the column with $y_i$ and $w_i$ so one possible matrix could be $$P_{BD}=\left( \begin{array}{ccc} -2 & 0 & 1 & 0\\ 1 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 \end{array} \right)$$ I checked that the columns are linearly independed so the matrix indeed represents a base change and if $D=\{d_1,d_2,d_3,d_4\}$ then each column represents $[d_i]_{B}.$ So $d_1=-1-3x+2x^2$, $d_2=1-x$, $d_3=1+x$ and $d_4=x^2-x^3.$ But this answer is incorrect. Any insights/hints will be much appreciated.
You need $d_1+2d_3=1-x+3x^2-x^3$ which is clearly not true, so your answer is wrong.
Your method seems unnecesarily complicated. You can satisfy the above equation by trial and error, say for example $$d_1=1-x+x^2-x^3\ ,\quad d_3=x^2\ ,$$ and then you just need two more vectors to complete a basis. May as well use standard basis vectors $1,x,x^2,x^3$ and throw out two of them. Clearly we throw out $x^2$ because we have it already. Work out which one of $1,x,x^3$ to throw out by row-reducing $$\pmatrix{1&0&1&0&0\cr -1&0&0&1&0\cr 1&1&0&0&0\cr -1&0&0&0&1\cr}\ .$$ Finally don't forget to put basis vectors in a suitable order ($d_3$ should come third not second). See if you can finish this.