I am reading the first page of this link http://math.mit.edu/~mckernan/Teaching/07-08/Spring/18.726/l_15.pdf.
This equality $$\left\{{h: R\to T: h \text{ homomorphism}} \right\} = \left\{{(a_1, ..., a_n): f_j (a_1, ..., a_n ) = 0, \forall {j}}\right\}$$ is what the paper means?
Thank you
I'm not entirely sure I understand your question, but indeed the claim in that PDF is correct, and it is what you described.
By the universal property of quotients, homomorphisms $R \to T$ are in bijective correspondence with homomorphisms $\varphi : \mathbb{Z}[x_1, \dots, x_n] \to T$ such that $\varphi(f_i) = 0$ for all $i$.
Since $\mathbb{Z}[x_1, \dots, x_n]$ is free on $\{x_1, \dots, x_n\}$, homomorphisms $\varphi : \mathbb{Z}[x_1, \dots, x_n] \to T$ are in bijective correspondence with tuples $(a_1, \dots, a_n) \in T^n$. Under this correspondence, the homomorphism determined by $(a_1, \dots, a_n)$ sends $f$ to $f(a_1, \dots, a_n)$.
So the set of homomorphisms $R \to T$ is therefore in bijective correspondence with the set of tuples $(a_1, \dots, a_n)$ such that $f_i(a_1, \dots, a_n) = 0$ for all $i$, exactly as you described. This bijection is natural in $T$.