The canonical comparison to the tensor functor of $F(A)$ is balanced and natural

For each $x\in X$ the map ${\mathrm{\ell }}_x:A\to X$, ${\mathrm{\ell }}_x(a)=ax$, is left $A$-linear, since $(a+a^{\mathrm{\prime }})x=ax+a^{\mathrm{\prime }}x$ and $(ca)x=c(ax)$; moreover ${\mathrm{\ell }}_{x+x^{\mathrm{\prime }}}={\mathrm{\ell }}_x+{\mathrm{\ell }}_{x^{\mathrm{\prime }}}$ and ${\mathrm{\ell }}_{ax}={\mathrm{\ell }}_x\circ r_a$, because ${\mathrm{\ell }}_{ax}(c)=(ca)x=c(ax)=({\mathrm{\ell }}_x\circ r_a)(c)$.

The pairing ${\beta }_X(m,x)=F({\mathrm{\ell }}_x)(m)$ is well defined because $F({\mathrm{\ell }}_x):M\to F(X)$ is a $B$-module homomorphism; it is additive in $m$ and $B$-linear in $m$ because $F({\mathrm{\ell }}_x)$ is, and additive in $x$ because $F({\mathrm{\ell }}_{x+x^{\mathrm{\prime }}})=F({\mathrm{\ell }}_x)+F({\mathrm{\ell }}_{x^{\mathrm{\prime }}})$ by additivity of $F$ and step 1.1. It is balanced: ${\beta }_X(ma,x)=F({\mathrm{\ell }}_x)(ma)=F({\mathrm{\ell }}_x)(F(r_a)(m))=F({\mathrm{\ell }}_x\circ r_a)(m)=F({\mathrm{\ell }}_{ax})(m)={\beta }_X(m,ax)$ by step 1.1, [F2] and functoriality.

By [F4] the balanced pairing ${\beta }_X$ induces a unique group homomorphism ${\tau }_X:M{\otimes }_{A}X\to F(X)$ with ${\tau }_X(m\otimes x)=F({\mathrm{\ell }}_x)(m)$. It is $B$-linear: by [F5] the left $B$-action on $M{\otimes }_{A}X$ satisfies $b(m\otimes x)=(bm)\otimes x$, and ${\tau }_X(b(m\otimes x))=F({\mathrm{\ell }}_x)(bm)=bF({\mathrm{\ell }}_x)(m)=b{\tau }_X(m\otimes x)$, because $F({\mathrm{\ell }}_x)$ is $B$-linear and elementary tensors generate; both sides are additive in the tensor variable, so equality on elementary tensors suffices.

Naturality: for left $A$-linear $u:X\to Y$ one has $u\circ {\mathrm{\ell }}_x={\mathrm{\ell }}_{u(x)}$, since $u(ax)=au(x)$, hence $F(u)\circ F({\mathrm{\ell }}_x)=F({\mathrm{\ell }}_{u(x)})$ by functoriality. Evaluating at $m$ and using $[F7]$ gives $F(u)({\tau }_X(m\otimes x))=F({\mathrm{\ell }}_{u(x)})(m)={\tau }_Y(m\otimes u(x))={\tau }_Y((1_M\otimes u)(m\otimes x))$; both sides are additive in the tensor variable and agree on elementary tensors, so $F(u)\circ {\tau }_X={\tau }_Y\circ (1_M\otimes u)$.

Steps 3.1 and 4.1 show that the components ${\tau }_X$ are $B$-linear maps assembling into a natural transformation $\tau :M{\otimes }_A-\Rightarrow F$ with ${\tau }_X(m\otimes x)=F({\mathrm{\ell }}_x)(m)$; the formulas used only the given element $x$ and the module $A$, never a presentation of $X$, and no element of an auxiliary set is selected, so no presentation and no choice are involved.