I have just read the definition of a flat module, and it is an $R$-module $F$ such that every monomorphism $M'\to M$ of $R$-modules induces a monomorphism: $$F\otimes M'\to F\otimes M,$$ but I wonder what induces means here.
I know that I have maps from $F,M'$ to $F\times M'$ and maps from $F,M$ to $F\times M$ as well as maps from $F\times M'\to F\otimes M'$ and $F\times M\to F\otimes M$, so I imagine my $M'\to M$ should induce our map via: $$\begin{array}{ccc}F\otimes M' &\stackrel{\psi^{\#}}\to&F\otimes M\\\uparrow&&\uparrow\\F\times M'&\stackrel{\psi}\to&F\times M\\\uparrow &&\uparrow\\M'&\longrightarrow&M \end{array}$$
Is that correct? So $\psi(f,m')=(a,m), \psi^\#(f\otimes m')=a\otimes m$, and we ask that the two squares commute?
The “induced morphism” by $f\colon M'\to M$ is the unique morphism $1\otimes f\colon F\otimes M'\to F\otimes M$ such that, for all $x\in F$ and $y\in M'$, $$ (1\otimes f)(x\otimes y)=x\otimes f(y) $$ By the properties of tensor products this is well defined.
It's essentially the upper square in your diagram. The lower square is not well defined, though: what should the map $M\to F\times M$ be?