Computing free resolution of $\dfrac{k[x,y,z,w]}{(x,y)\cap(z,w)}$ as $k[x,y,z,w]$-module This is a question related to Hilbert polynomial of disjoint union of lines in $\Bbb{P}^3$. 
It seems everyone finds computation of the free resolution quite simple here. I do not see how to compute $\phi_1$ so easily. 
Let $R=k[x,y,z,w]$, $S(X)=\dfrac{k[x,y,z,w]}{(x,y)\cap (z,w)}$ and $R\to S\to 0$ as $R$-module.
I knew $R(-2)^4\xrightarrow{\phi_2} \ker(R\to S(X))\to 0$ for sure. 
The question would be how to compute the kernel of $\phi_2$.
Let $e_1,\dots, e_4$ be a basis of $R(-2)^4$ free module. The map sends $\phi_2(e_1)=xy,\phi_2(e_2)=xw,\phi_2(e_3)=yz,\phi_2(e_4)=yw$. My approach is to first check $(we_1-ze_2,ye_1-xe_3,ye_2-xe_4,we_3-ze_4)\subset \ker(\phi_2)$.
Then identify $$\frac{R(-2)^4}{(we_1-ze_2,ye_1-xe_3,ye_2-xe_4,we_3-ze_4)}=k[x,z]e_1\oplus k[x,z,w]e_2\oplus k[x,y,z]e_3\oplus k[x,y,z,w]e_4,$$ where the latter is all in class notation.
Then write out a representative of an element to check induced map of above quotient to $\ker(\phi_2)$ is injective by domain property as $\ker(\phi_2)\subset R$. Surjectivity is obvious. Then I can set up $\phi_2$ as in the post and keep working on the rest. 
Q: Is there a less painful way to deal with finding free resolution? Obviously Hilbert syzygy theorem only asserts there is a free resolution of finite length for this problem.
 A: I will try to explain how to obtain the maps in the answer you link to.
It is easier to begin with the fact $(x,y)\cap (z,w) = (xz,xw,yz,yw)$. Then your module $M$ is a quotient of $R = k[x,y,z,w]$ by such ideal $I$. There is first, the standard projection $R\to M$. This has kernel the ideal $I$ which can be covered by four copies of $R$, one for each relation, to obtain a sequence
$$R(-2)^4\to R\to M\to 0$$
Since the ideal has relations in degree $2$, this shifts the degree by such a number. Now you need to find the kernel of the map $$(p_1,p_2,p_3,p_4)\longmapsto xwp_1+xzp_2+ywp_3+yzp_4 $$
But from $xwp_1+xzp_2+ywp_3+yzp_4=0$ you can deduce, essentially by divisibility considerations, that there exist $(q_1,q_2,q_3,q_4)$ such that $$p_1 = \hphantom{+}zq_1+yq_2 \\p_2 = -wq_1+yq_3\\ p_3=-xq_3+wq_4\\p_4 = -xq_2-zq_4.$$ 
Since again the variables have degree $1$, this defines a map $R(-1)^4\to R(-2)^4$ given by the matrix in the answer of the link. It is not complicated to find the kernel of this matrix now since 
$$0 = \hphantom{+}zq_1+yq_2 \\0 = -wq_1+yq_3\\ 0=-xq_3+wq_4\\0= -xq_2-zq_4.$$ 
entail again by divisibility that $(q_1,q_2,q_3,q_4) = (yr,-rz,rw,-rx)$ for some polynomial $r$. Clearly this is injective and thus completes the resolution.
Add Suppose that $xwp_1+xzp_2+ywp_3+yzp_4=0$. Since the last two terms are in $(y)$ and since $x\notin (y)$, it follows that $wp_1+zp_2 = yf_{12}$ for some polynomial $f_{12}$. Similarly
$$wp_1+zp_2 = yf_{12}\\
xp_1+yp_3 = zf_{13}\\
xp_2+yp_4 = wf_{24}\\
wp_3+zp_4 = xf_{34}$$
Then we see that $xyf_{12}+yxf_{34}=0$ so $f_{12} = -f_{34}$. Similarly $f_{13} = - f_{24}$, so we get 
$$wp_1+zp_2 = yf_{12}\\
xp_1+yp_3 = zf_{13}\\
xp_2+yp_4 = -wf_{13}\\
wp_3+zp_4 = -xf_{12}$$
Now since $x,y\notin (w,z)$ and $w,z\notin (x,y)$ we get that $f_{12} = wa_1+za_2$ and $f_{13} = xa_3+ya_4$. This means that 
$$w(p_1-ya_1) + z(p_2-ya_2) = 0 $$
so that $p_1 = ya_1 + zb_2$ and $p_2 = ya_2 - wb_2$. Continuing in this way for all other equations gives the above. 
