First we prove (1) implies (2).  Suppose $M$ is Mittag-Leffler and let $x$ be in the kernel of $M \otimes_R (\prod_{\alpha} Q_{\alpha}) \to \prod_{\alpha} (M \otimes_R Q_{\alpha})$.  Write $M$ as a colimit $M = \colim_{i \in I} M_i$ of a directed system of finitely presented modules $M_i$.  Then $M \otimes_R (\prod_{\alpha} Q_{\alpha})$ is the colimit of $M_i \otimes_R (\prod_{\alpha} Q_{\alpha})$.  So $x$ is the image of an element $x_i \in M_i \otimes_R (\prod_{\alpha} Q_{\alpha})$.  We must show that $x_i$ maps to $0$ in $M_j \otimes_R (\prod_{\alpha} Q_{\alpha})$ for some $j \geq i$.  Since $M$ is Mittag-Leffler, we may choose $j \geq i$ such that $M_i \to M_j$ and $M_i \to M$ dominate each other.  Then consider the commutative diagram $$ \xymatrix{ M \otimes_R (\prod_{\alpha} Q_{\alpha}) \ar[r] & \prod_{\alpha} (M \otimes_R Q_{\alpha}) \\ M_i \otimes_R (\prod_{\alpha} Q_{\alpha}) \ar[r]^{\cong} \ar[d] \ar[u] & \prod_{\alpha} (M_i \otimes_R Q_{\alpha}) \ar[d] \ar[u] \\ M_j \otimes_R (\prod_{\alpha} Q_{\alpha}) \ar[r]^{\cong} & \prod_{\alpha} (M_j \otimes_R Q_{\alpha}) } $$ whose bottom two horizontal maps are isomorphisms, according to Proposition \ref{proposition-fp-tensor}.  Since $x_i$ maps to $0$ in $\prod_{\alpha} (M \otimes_R Q_{\alpha})$, its image in $\prod_{\alpha} (M_i \otimes_R Q_{\alpha})$ is in the kernel of the map $\prod_{\alpha} (M_i \otimes_R Q_{\alpha}) \to \prod_{\alpha} (M \otimes_R Q_{\alpha})$.  But this kernel equals the kernel of $\prod_{\alpha} (M_i \otimes_R Q_{\alpha}) \to \prod_{\alpha} (M_j \otimes_R Q_{\alpha})$ according to the choice of $j$.  Thus $x_i$ maps to $0$ in $\prod_{\alpha} (M_j \otimes_R Q_{\alpha})$ and hence to $0$ in $M_j \otimes_R (\prod_{\alpha} Q_{\alpha})$. \medskip\noindent Now suppose (2) holds. We prove $M$ satisfies formulation (1) of being Mittag-Leffler from Proposition \ref{proposition-ML-characterization}.  Let $f: P \to M$ be a map from a finitely presented module $P$ to $M$.  Choose a set $B$ of representatives of the isomorphism classes of finitely presented $R$-modules. Let $A$ be the set of pairs $(Q, x)$ where $Q \in B$ and $x \in \Ker(P \otimes Q \to M \otimes Q)$.  For $\alpha = (Q, x) \in A$, we write $Q_{\alpha}$ for $Q$ and $x_{\alpha}$ for $x$.  Consider the commutative diagram $$ \xymatrix{ M \otimes_R (\prod_{\alpha} Q_{\alpha}) \ar[r] & \prod_{\alpha} (M \otimes_R Q_{\alpha}) \\ P \otimes_R (\prod_{\alpha} Q_{\alpha}) \ar[r]^{\cong} \ar[u] & \prod_{\alpha} (P \otimes_R Q_{\alpha}) \ar[u] . } $$ The top arrow is an injection by assumption, and the bottom arrow is an isomorphism by Proposition \ref{proposition-fp-tensor}.  Let $x \in P \otimes_R (\prod_{\alpha} Q_{\alpha})$ be the element corresponding to $(x_{\alpha}) \in \prod_{\alpha} (P \otimes_R Q_{\alpha})$ under this isomorphism.  Then $x \in \Ker( P \otimes_R (\prod_{\alpha} Q_{\alpha}) \to M \otimes_R (\prod_{\alpha} Q_{\alpha}))$ since the top arrow in the diagram is injective.  By Lemma \ref{lemma-kernel-tensored-fp}, we get a finitely presented module $P'$ and a map $f': P \to P'$ such that $f: P \to M$ factors through $f'$ and $x \in \Ker(P \otimes_R (\prod_{\alpha} Q_{\alpha}) \to P' \otimes_R (\prod_{\alpha} Q_{\alpha}))$.  We have a commutative diagram $$ \xymatrix{ P' \otimes_R (\prod_{\alpha} Q_{\alpha}) \ar[r]^{\cong} & \prod_{\alpha} (P' \otimes_R Q_{\alpha}) \\ P \otimes_R (\prod_{\alpha} Q_{\alpha}) \ar[r]^{\cong} \ar[u] & \prod_{\alpha} (P \otimes_R Q_{\alpha}) \ar[u] . } $$ where both the top and bottom arrows are isomorphisms by Proposition \ref{proposition-fp-tensor}.  Thus since $x$ is in the kernel of the left vertical map, $(x_{\alpha})$ is in the kernel of the right vertical map.  This means $x_{\alpha} \in \Ker(P \otimes_R Q_{\alpha} \to P' \otimes_R Q_{\alpha})$ for every $\alpha \in A$.  By the definition of $A$ this means $\Ker(P \otimes_R Q \to P' \otimes_R Q) \supset \Ker(P \otimes_R Q \to M \otimes_R Q)$ for all finitely presented $Q$ and, since $f: P \to M$ factors through $f': P \to P'$, actually equality holds.  By Lemma \ref{lemma-domination-fp}, $f$ and $f'$ dominate each other.