multiplicative structure of Ext

So, maybe I should format this as an answer.

Consider any pair of objects $A, B$ in an abelian category with, say, enough projective objects. Then, the extension group $Ext^i(A,B)$ is defined as $H^i(Hom(A^{\bullet}, B^{\bullet}))$, where $A^{\bullet}, B^{\bullet}$ are resolutions of objects $A, B$ respectively, and $A^{\bullet}$ needs to be projective, and no requirement on $B^{\bullet}$ but we will take it to be projective too to define composition.

Composition for Exts is then induced by composition $Hom(A^{\bullet}, B^{\bullet}) \otimes Hom(B^{\bullet}, C^{\bullet}) \rightarrow Hom(A^{\bullet}, C^{\bullet})$.

Let us denote $R = \mathbb{F}_2[x]/x^2$. Then, $\mathbb{F}_2$ admits the following resolution: $A^{\bullet} = ... \rightarrow \Sigma^2 R \rightarrow \Sigma R \rightarrow R \rightarrow 0$, the maps being multiplication by $x$.

The hom-complex $Hom(A^{\bullet},A^{\bullet})$ is then quasiisomorphic to $Hom(A^{\bullet}, \mathbb{F}_2)$ (it is always true for any pair of complexes such that left one is a complex of projective modules and the right one is switched to the quasiisomorphic one). Concretely, it means that any closed map $\phi: A^{\bullet} \rightarrow \mathbb{F}_2$ can be lifted to the map $\phi: A^{\bullet} \rightarrow A^{\bullet}$ uniquely up to homotopy. So, the map we lift is your $\phi_t$, and the lift is defined as $\phi_t: A^k \rightarrow A^{k-t} = Id$ for $k \geq t$ and $0$ otherwise.

It is now clear that $\phi_t \phi_s = \phi_{t+s}$

$\blacksquare$

the rule is basically "switch everything to resolutions and then map complexes instead of objects". it works very well if you have a category with enough projective or injective objects

The composition product $Ext(N,P) \otimes Ext(M,N) \to Ext(M,P)$ can be computed as follows. Given cocycles $x : N_{s_1} \to \Sigma^{t_1} P$ and $y : M_{s_2} \to \Sigma^{t_2} N$, let $\{ y_s : M_{s+s_2} \to \Sigma^{t_2} N_s\}$ be a chain map lifting $y$. Then $xy$ is represented by the cocycle $x y_{s_1} : M_{s_1+s_2} \to \Sigma^{t_2} N_{s_1} \to \Sigma^{t_1+t_2} P $.

The point is that one needs (a finite bit of) the chain map lifting $y$, but only the cocycle $x$.

Saves effort.

(The $M_s$ and $N_s$ are the terms in projective resolutions of $M$ and $N$, of course.)

In this case, you're dealing with a Koszul algebra ---this is precisely your observation that the spectral sequence is concentrated in the diagonal. Its Koszul dual is a polynomial algebra, and this gives you the multiplicative structure on $\mathsf{Ext}$: each diagonal entry is one dimensional with generator $x_s$, and you have that $x_sx_t=x_{t+s}$, so $\mathsf{Ext}$ is a polynonimal algebra generated by $x_1$.

This can be seen at the level of the (reduced) bar construction $B\Gamma$, whose dual computes your $E_2$-page. It has zero differential, and it is generated in each homological degree $s$ by $x_s = [x\vert\cdots\vert x]$ with $x$ appearing $s$ times. The product is dual to the coproduct of $B\Gamma$, which is simply $\Delta(x_s) = \sum x_i\otimes x_j$.