Symmetric algebra of monomial ideal modulo a binomial ideal isomorphic to a quotient of a polynomial ring

Question

Note: I posted this question on MSE but haven't received any response.

I’m trying to understand the proof of Corollary 1.9 in “Binomial ideals” by David Eisenbud and Bernd Sturmfels.

Notation: Let $S= k[x_1, \ldots, x_n]$ where $k$ is a field and the $x_i$ are indeterminants, $M = (m_1, \ldots, m_t)$ a monomial ideal in $S$, and $B$ a binomial ideal in $S$. Set $R=S/B$ and $I = (B+M)/B \subseteq R$.

The authors say that we may represent the symmetric algebra of $I$ over $R$, denoted $Sym_R I$, as a quotient of a polynomial ring $R[y_1, \ldots, y_t]$,

I’m trying to understand this isomorphism. Define the map $\phi: R[y_1, \ldots, y_t] \rightarrow Sym_RI$ with $y_i \mapsto m_i$. Obviously the relations $\sum_j f_{i,j}y_j$ are in the kernel of $\phi$ since they represent syzygies of $I$. But how do we know that the kernel consists only of these relations?

For example, suppose that $g_1y_1^2 + g_2y_1y_2 + g_3y_2y_3 \mapsto g_1(m_1 \odot m_1) + g_2(m_1 \odot m_2) + g_3(m_2 \odot m_3) = 0$, where $\odot$ repserents the product in the symmetric algebra. Then $g_1y_1^2 + g_2y_1y_2 + g_3y_2y_3$ is in the kernel but I can’t see any way to express it in the form $\sum_j f_{i,j}y_j$