Suppose that $\psi\colon L\to M$ is an injective homomorphism of $R$-modules and $F\cong R^n$. I would like to show that $1\otimes \psi\colon F\otimes_R L\to F\otimes_R M$ is also injective (this is a step in the proof that free module are flat). I know that $F\otimes_R L = R^n\otimes_R L \cong L^n$ and likewise $F\otimes_R M \cong M^n$. According to D&F page 401,
under these isomorphisms the map $1\otimes \psi\colon F\otimes_R L\to F\otimes_R M$ is just the natural map of $L^n$ to $M^n$ induced by the inclusion $\psi$ in each component.
I'm having a hard time seeing why $1\otimes\psi$ is the map of $L^n$ to $M^n$ induced by the inclusion $\psi$ in each component. From what I understand, $1\otimes\psi$ is the unique $R$-module homomorphism satisfying $$ (1\otimes\psi)(f\otimes l) = f\otimes \psi(l). $$ Since $F\cong R^n$ I guess I could write the above map as $$ (1\otimes\psi)((r_1,\dots,r_n)\otimes l) = (r_1,\dots,r_n)\otimes \psi(l). $$ But I'm still not sure where the fact that $1\otimes\psi$ is injective is coming from. (Btw, I haven't learned category theory yet, so I'm trying to see this without appealing to ideas from category theory.)
Regarding how the map $1 \otimes \psi$ acts on $F \otimes L \cong L^n$, it helps to think about the specific isomorphism $F \otimes L \overset{\omega}{\to} L^n$, in order to make it clear to yourself that the following diagram commutes: $$\require{AMScd} \begin{CD} F \otimes L @>{1 \otimes \psi}>> F \otimes M \\ @V{\omega}V{\sim}V @V{\omega}V{\sim}V \\ L^n @>{\psi^n}>> M^n \end{CD}$$ You can read about how to think about $\omega$ here on page 17 of Conrad's notes, for instance.
As for why the map is injective, it shouldn't be too hard to convince yourself that a map $L^n \to M^n$ which is just an $n$-fold product of injective maps is itself injective. (There's no more trickiness coming from the tensor products now.) Because the sides of the above diagram are isomorphisms, injectivity of $\psi^n$ implies injectivity of $1 \otimes \psi$.