I know that for a flat $R$-module $M$ the property beeing fathfully flat is in most literature is characterized via
(*) for every complex of $R$-modules $N_1→N_2→N_3$ is exact if and only if the sequence $M⊗_R N_1→M⊗_ RN_2→M⊗_R N_3$ is exact.
I would like to see why and how to see that- if we assume that $M$ is a $R$-algebra(!) - faithfully flatness can be equivalently characterized by the condition that the corresponding map between Spectra
(**) $$\operatorname{Spec}(A) \to \operatorname{Spec}(R)$$ is surjective?
This is a classical result of commutative algebra, which can be found in various books, such as the Corollary 2.20 in Liu's Algebraic Geometry and Arithmetic Curves, the Exercise 3.16 of Atiyah & MacDonald's Introduction to Commutative Algebra and the Theorem 3 in Chap.1 of Matsumura's Commutative Algebra. Suppose that the ring homomorphism $f\colon A\to B$ is flat. $\newcommand{\m}{\mathfrak m}$ $\newcommand{\spec}{\operatorname{Spec}}$ $\newcommand{\p}{\mathfrak p}$ To see this, you may need to know firstly these results about faithful flatness:
$f$ is faithfully flat iff for each $A$-module $M\neq 0$, $M\otimes_A B\neq 0$.
$f$ is faithfully flat iff $\m B\neq B$ for any maximal ideal $\m$.
The following proof is essentially an excerpt of Matsumura's book. Let $f\colon A\to B$ be a flat ring homomorphism. Then
$f$ is faithfully flat $\Longrightarrow$ $f^*\colon \spec(B)\to\spec(A)$ is surjective. To this end, for any $\p\in\spec(A)$, we have $\p B_\p\neq B_\p$, where $B_\p=A_\p\otimes_A B$ (otherwise $0=B_\p/\p B_\p=(A_\p/ \p A_\p)\otimes_A B\neq 0$, a contradiction). Pick a maximal ideal $\m\supset\p B_\p$ of $B_\p$. Since the following diagram is commutative $$\require{AMScd} \begin{CD} A@>{f}>> B\\ @V{\varphi_A}VV@VV{\varphi_B=\varphi_A\otimes \mathrm{id}_B}V\\ A_\p@>{f_\p}>> B_\p\end{CD}$$where $\varphi_A\colon A\to A_\p$ is the localization map, it suffices to take $P=\varphi_B^{-1}(\m)$. Indeed, since $\p A_\p\subset f_\p^{-1}(\m)$ and $\p A_\p$ is a maximal ideal, $f_\p^{-1}(\m)=\p A_\p$. Then it follows that$$ f^*(P)=f^{-1}(P)=f^{-1}(\varphi_B^{-1}(\m))=\varphi_A^{-1}(f_\p^{-1}(\m))=\varphi_A^{-1}(\p A_\p)=\p. $$
$f^*$ is surjective $\Longrightarrow$ $f$ is faithfully flat. For each maximal ideal $\m$ of $A$, there is a $P\in\spec(B)$ such that $f^*(P)=\m$. Thus for any maximal ideal $\m'\supset P$ of $B$, we have $f^*(\m')=\m$, i.e., $\m'$ lies over $\m$, which implies $\m B\neq B$. The arbitrariness of $\m$ entails the faithful flatness of $f$.