Obviously (1) implies (2). \medskip\noindent Next we show (2) implies (3).  Let $f_1(x_i) = \sum_j a_{ij} y_j$ be relations as in (3).  Let $(d_j)$ be a basis for $R^m$, $(e_i)$ a basis for $R^n$, and $R^m \to R^n$ the map given by $d_j \mapsto \sum_i a_{ij} e_i$. Let $Q$ be the cokernel of $R^m \to R^n$.  Then tensoring $R^m \to R^n \to Q \to 0$ by the map $f_1: M_1 \to M_2$, we get a commutative diagram $$ \xymatrix{ M_1^{\oplus m} \ar[r] \ar[d] & M_1^{\oplus n} \ar[r] \ar[d] & M_1 \otimes_R Q \ar[r] \ar[d] & 0 \\ M_2^{\oplus m} \ar[r] & M_2^{\oplus n} \ar[r] & M_2 \otimes_R Q \ar[r] & 0 } $$ where $M_1^{\oplus m} \to M_1^{\oplus n}$ is given by $$ (z_1, \ldots, z_m) \mapsto (\sum\nolimits_j a_{1j} z_j, \ldots, \sum\nolimits_j a_{nj} z_j), $$ and $M_2^{\oplus m} \to M_2^{\oplus n}$ is given similarly.  We want to show $x = (x_1, \ldots, x_n) \in M_1^{\oplus n}$ is in the image of $M_1^{\oplus m} \to M_1^{\oplus n}$.  By (2) the map $M_1 \otimes Q \to M_2 \otimes Q$ is injective, hence by exactness of the top row it is enough to show $x$ maps to $0$ in $M_2 \otimes Q$, and so by exactness of the bottom row it is enough to show the image of $x$ in $M_2^{\oplus n}$ is in the image of $M_2^{\oplus m} \to M_2^{\oplus n}$.  This is true by assumption. \medskip\noindent Condition (4) is just a translation of (3) into diagram form. \medskip\noindent Next we show (4) implies (5). Let $\varphi : P \to M_3$ be a map from a finitely presented $R$-module $P$.  We must show that $\varphi$ lifts to a map $P \to M_2$.  Choose a presentation of $P$, $$ R^n \xrightarrow{g_1} R^m \xrightarrow{g_2} P \to 0. $$ Using freeness of $R^n$ and $R^m$, we can construct $h_2: R^m \to M_2$ and then $h_1: R^n \to M_1$ such that the following diagram commutes $$ \xymatrix{          & R^n \ar[r]^{g_1} \ar[d]^{h_1} & R^m \ar[r]^{g_2} \ar[d]^{h_2} & P \ar[r] \ar[d]^{\varphi} & 0 \\ 0 \ar[r] & M_1 \ar[r]^{f_1} & M_2 \ar[r]^{f_2} & M_3 \ar[r] & 0 . } $$ By (4) there is a map $k_1: R^m \to M_1$ such that $k_1 \circ g_1 = h_1$.  Now define $h'_2: R^m \to M_2$ by $h_2' = h_2 - f_1 \circ k_1$. Then $$ h'_2 \circ g_1 = h_2 \circ g_1 - f_1 \circ k_1 \circ g_1 = h_2 \circ g_1 - f_1 \circ h_1 = 0 . $$ Hence by passing to the quotient $h'_2$ defines a map $\varphi': P \to M_2$ such that $\varphi' \circ g_2 = h_2'$.  In a diagram, we have $$ \xymatrix{ R^m \ar[r]^{g_2} \ar[d]_{h'_2} & P \ar[d]^{\varphi} \ar[dl]_{\varphi'} \\ M_2 \ar[r]^{f_2} & M_3. } $$ where the top triangle commutes.  We claim that $\varphi'$ is the desired lift, i.e.\ that $f_2 \circ \varphi' = \varphi$.  From the definitions we have $$ f_2 \circ \varphi' \circ g_2 = f_2 \circ h'_2 = f_2 \circ h_2 - f_2 \circ f_1 \circ k_1 = f_2 \circ h_2 = \varphi \circ g_2. $$ Since $g_2$ is surjective, this finishes the proof. \medskip\noindent Now we show (5) implies (6). Write $M_{3}$ as the colimit of a directed system of finitely presented modules $M_{3, i}$, see Lemma \ref{lemma-module-colimit-fp}. Let $M_{2, i}$ be the fiber product of $M_{3, i}$ and $M_{2}$ over $M_{3}$---by definition this is the submodule of $M_2 \times M_{3, i}$ consisting of elements whose two projections onto $M_3$ are equal. Let $M_{1, i}$ be the kernel of the projection $M_{2, i} \to M_{3, i}$. Then we have a directed system of exact sequences $$ 0 \to M_{1, i} \to M_{2, i} \to M_{3, i} \to 0, $$ and for each $i$ a map of exact sequences $$ \xymatrix{ 0  \ar[r] & M_{1, i} \ar[d] \ar[r]  &  M_{2, i}  \ar[r] \ar[d] & M_{3, i} \ar[d] \ar[r] & 0 \\ 0  \ar[r] & M_{1} \ar[r]  &  M_{2}  \ar[r] & M_{3} \ar[r] & 0 } $$ compatible with the directed system.  From the definition of the fiber product $M_{2, i}$, it follows that the map $M_{1, i} \to M_1$ is an isomorphism.  By (5) there is a map $M_{3, i} \to M_{2}$ lifting $M_{3, i} \to M_3$, and by the universal property of the fiber product this gives rise to a section of $M_{2, i} \to M_{3, i}$.  Hence the sequences $$ 0 \to M_{1, i} \to M_{2, i} \to M_{3, i} \to 0 $$ split.  Passing to the colimit, we have a commutative diagram $$ \xymatrix{ 0  \ar[r] & \colim M_{1, i} \ar[d]^{\cong} \ar[r]  &  \colim M_{2, i}  \ar[r] \ar[d] & \colim M_{3, i} \ar[d]^{\cong} \ar[r] & 0 \\ 0  \ar[r] & M_{1} \ar[r]  &  M_{2}  \ar[r] & M_{3} \ar[r] & 0 } $$ with exact rows and outer vertical maps isomorphisms.  Hence $\colim M_{2, i} \to M_2$ is also an isomorphism and (6) holds. \medskip\noindent Condition (6) implies (1) by Example \ref{example-universally-exact} (2).