associated graded polynomial surjection

The degree-zero part is $R\mathrm{/}\mathfrak{m}=k$. On ${\mathfrak{m}}^n\mathrm{/}{\mathfrak{m}}^{n+1}$ the action of $R$ factors through $k$, since $\mathfrak{m}{\mathfrak{m}}^n\subseteq {\mathfrak{m}}^{n+1}$. The graded multiplication therefore defines the displayed polynomial map. Altering a representative by ${\mathfrak{m}}^{n+1}$ changes a product of degrees $n,j$ by ${\mathfrak{m}}^{n+j+1}$, so multiplication and the map are well-defined.

For $n\ge 1$, every element of ${\mathfrak{m}}^n\mathrm{/}{\mathfrak{m}}^{n+1}$ is a sum of classes of products $u_1\cdots u_n$ with each $u_j\in \mathfrak{m}$. In $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, express the class of each $u_j$ as a $k$-linear combination of the given basis classes $x_i+{\mathfrak{m}}^2$. Replacing one factor $u_j$ by such a combination changes the product only by an element of ${\mathfrak{m}}^{n+1}$. Expanding therefore expresses every degree-$n$ class as a $k$-linear combination of monomials in the $x_i+{\mathfrak{m}}^2$, all in the image of $\varphi$. Degree zero is $k$ by step 1.1. This argument also covers $e=0$: the empty degree-one basis makes every positive-degree quotient zero, so the map $k\to {\mathrm{gr}}_{\mathfrak{m}}R$ is surjective.