First suppose $M$ is finite projective, i.e., (2) holds. Take a surjection $R^n \to M$ and let $K$ be the kernel. Since $M$ is projective, $0 \to K \to R^n \to M \to 0$ splits. Hence (2) $\Rightarrow$ (3). The implication (3) $\Rightarrow$ (2) follows from the fact that a direct summand of a projective is projective, see Lemma \ref{lemma-characterize-projective}. \medskip\noindent Assume (3), so we can write $K \oplus M \cong R^{\oplus n}$. So $K$ is a  direct summand of $R^n$ and thus finitely generated. This shows $M = R^{\oplus n}/K$ is finitely presented. In other words, (3) $\Rightarrow$ (1). \medskip\noindent Assume $M$ is finitely presented and flat, i.e., (1) holds. We will prove that (7) holds. Pick any prime $\mathfrak p$ and $x_1, \ldots, x_r \in M$ which map to a basis of $M \otimes_R \kappa(\mathfrak p)$. By Nakayama's Lemma \ref{lemma-NAK} these elements generate $M_g$ for some $g \in R$, $g \not \in \mathfrak p$. The corresponding surjection $\varphi : R_g^{\oplus r} \to M_g$ has the following two properties: (a) $\Ker(\varphi)$ is a finite $R_g$-module (see Lemma \ref{lemma-extension}) and (b) $\Ker(\varphi) \otimes \kappa(\mathfrak p) = 0$ by flatness of $M_g$ over $R_g$ (see Lemma \ref{lemma-flat-tor-zero}). Hence by Nakayama's lemma again there exists a $g' \in R_g$ such that $\Ker(\varphi)_{g'} = 0$. In other words, $M_{gg'}$ is free. \medskip\noindent A finite locally free module is a finite module, see Lemma \ref{lemma-cover}, hence (7) $\Rightarrow$ (6). It is clear that (6) $\Rightarrow$ (7) and that (7) $\Rightarrow$ (8). \medskip\noindent A finite locally free module is a finitely presented module, see Lemma \ref{lemma-cover}, hence (7) $\Rightarrow$ (4). Of course (4) implies (5). Since we may check flatness locally (see Lemma \ref{lemma-flat-localization}) we conclude that (5) implies (1). At this point we have $$ \xymatrix{ (2) \ar@{<=>}[r] & (3) \ar@{=>}[r] & (1) \ar@{=>}[r] & (7)  \ar@{<=>}[r] \ar@{=>}[rd] \ar@{=>}[d] & (6) \\ & & (5) \ar@{=>}[u] & (4) \ar@{=>}[l] & (8) } $$ \medskip\noindent Suppose that $M$ satisfies (1), (4), (5), (6), and (7). We will prove that (3) holds. It suffices to show that $M$ is projective. We have to show that $\Hom_R(M, -)$ is exact. Let $0 \to N'' \to N \to N'\to 0$ be a short exact sequence of $R$-module. We have to show that $0 \to \Hom_R(M, N'') \to \Hom_R(M, N) \to \Hom_R(M, N') \to 0$ is exact. As $M$ is finite locally free there exist a covering $\Spec(R) = \bigcup D(f_i)$ such that $M_{f_i}$ is finite free. By Lemma \ref{lemma-hom-from-finitely-presented} we see that $$ 0 \to \Hom_R(M, N'')_{f_i} \to \Hom_R(M, N)_{f_i} \to \Hom_R(M, N')_{f_i} \to 0 $$ is equal to $0 \to \Hom_{R_{f_i}}(M_{f_i}, N''_{f_i}) \to \Hom_{R_{f_i}}(M_{f_i}, N_{f_i}) \to \Hom_{R_{f_i}}(M_{f_i}, N'_{f_i}) \to 0$ which is exact as $M_{f_i}$ is free and as the localization $0 \to N''_{f_i} \to N_{f_i} \to N'_{f_i} \to 0$ is exact (as localization is exact). Whence we see that $0 \to \Hom_R(M, N'') \to \Hom_R(M, N) \to \Hom_R(M, N') \to 0$ is exact by Lemma \ref{lemma-cover}. \medskip\noindent Finally, assume that (8) holds. Pick a maximal ideal $\mathfrak m \subset R$. Pick $x_1, \ldots, x_r \in M$ which map to a $\kappa(\mathfrak m)$-basis of $M \otimes_R \kappa(\mathfrak m) = M/\mathfrak mM$. In particular $\rho_M(\mathfrak m) = r$. By Nakayama's Lemma \ref{lemma-NAK} there exists an $f \in R$, $f \not \in \mathfrak m$ such that $x_1, \ldots, x_r$ generate $M_f$ over $R_f$. By the assumption that $\rho_M$ is locally constant there exists a $g \in R$, $g \not \in \mathfrak m$ such that $\rho_M$ is constant equal to $r$ on $D(g)$. We claim that $$ \Psi : R_{fg}^{\oplus r} \longrightarrow M_{fg}, \quad (a_1, \ldots, a_r) \longmapsto \sum a_i x_i $$ is an isomorphism. This claim will show that $M$ is finite locally free, i.e., that (7) holds. To see the claim it suffices to show that the induced map on localizations $\Psi_{\mathfrak p} : R_{\mathfrak p}^{\oplus r} \to M_{\mathfrak p}$ is an isomorphism for all $\mathfrak p \in D(fg)$, see Lemma \ref{lemma-characterize-zero-local}. By our choice of $f$ the map $\Psi_{\mathfrak p}$ is surjective. By assumption (8) we have $M_{\mathfrak p} \cong R_{\mathfrak p}^{\oplus \rho_M(\mathfrak p)}$ and by our choice of $g$ we have $\rho_M(\mathfrak p) = r$. Hence $\Psi_{\mathfrak p}$ determines a surjection $R_{\mathfrak p}^{\oplus r} \to M_{\mathfrak p} \cong R_{\mathfrak p}^{\oplus r}$ whence is an isomorphism by Lemma \ref{lemma-fun}. (Of course this last fact follows from a simple matrix argument also.)