Conditions for null-homotopic chain map in a specific example

Approach 1: minimal resolution: consider a building a resolution of $\mathbb k$ by hand, which you can choose to be a periodic Koszul resolution, as you did:

$$\xrightarrow{\phantom{m}\cdots\phantom{m}} R^2 \xrightarrow{\left[\begin{matrix} 0 & y\\x & 0 \end{matrix} \right]} R^2 \xrightarrow{\left[\begin{matrix} 0 & y\\x & 0 \end{matrix} \right]} R^2 \xrightarrow{\left[\begin{matrix} x \\y \end{matrix} \right]} R \longrightarrow 0 $$

This resolution is minimal, in the sense that taking $\mathrm{Hom}$ with $\mathbb k$ yields a zero differential. Since the resulting complex is just

$$\xrightarrow{\phantom{m}0\phantom{m}} \mathbb k^2 \xrightarrow{\phantom{m}0\phantom{m}} \mathbb k^2 \xrightarrow{\phantom{m}0\phantom{m}} \mathbb k^2 \xrightarrow{\phantom{m}0\phantom{m}} \mathbb k\longrightarrow 0 $$

this immediately tells you that $\mathrm{Ext}_R^n(\mathbb k, \mathbb k)$ is two dimensional for all $n\geqslant 2$. Moreover, you can show that the augmentation $\varepsilon:P\longrightarrow \mathbb k$ induces a quasi-iso $\varepsilon_*:\mathrm{End}(P) \to \mathrm{Hom}(P,\mathbb k)$, so you don't have to compute the homology of the endomorphism complex by hand.

Approach 2: Koszul duals. There is a simpler approach to computing $\mathrm{Ext}$ in this case.

Namely, if you consider $A = \mathbb k[x,y]/(xy)$, then this is a monomial associative algebra presented by $x$ and $y$ modulo two relations $xy=0$ and $yx=0$ (note that $xy-yx=0$ can be replaced by the second once since $xy=0$). That is, $A= \mathbb k\langle x,y\rangle / (xy,yx)$.

In this case, the normal forms (things not divisible by $xy$ or $yx$) are $1$ and $x^n,y^n$ for $n\geqslant 1$, so the kernel of multiplication by $x$ is the ideal generated by $y$, and the kernel of multiplication by $y$ is the ideal generated by $x$.

Having said this, quadratic monomial algebras are Koszul, the Koszul dual in this case is $\mathbb k\langle x,y\rangle/(x^2,y^2)$, since all binomials are $xy,yx,y^2,x^2$, and we merely need to pick those not appearing in the presentation of $A$. Hence, $\mathrm{Ext}_A(\mathbb k,\mathbb k)$ as an algebra is isomorphic to this algebra.

Approach 3: Anick chains: For monomial algebras, there is a completely combinatorial way to compute the Yoneda (co)algebra. In this case, the monomial relations are $xy$ and $yx$, and the $n$-chains are equal to the two element set $$C_n = \{ X_n = x(yx)^{n-1}, Y_n = y(xy)^{n-1}\}$$ for all $n\geqslant 1$. The Anick resolution then is of the form $(R\otimes C_\bullet ,d)$ so that $R\otimes C_n = R^2$ for all $n\geqslant 1$ and the differential is $dX_n = x\otimes Y_{n-1}$ and $dY_n = y\otimes X_{n-1}$, i.e. this is isomorphic to the resolution you found, and the resolution given by Koszul duality (because one is dealing with quadratic monomial algebras here).