This is related to base change theorem 4.3.7 of Weibel, Homological Algebra of Chpt 4, Sec 3. $\DeclareMathOperator{\gldim}{gl-dim}$Denote $\gldim$ as global dimension of ring $R$ and assume $\gldim(R)<\infty$.
Thm(special case) 4.3.7 If $R[x]$ is a polynomial ring, then $\gldim(R[x])=1+\gldim(R)$.
One of the step is to show $\gldim(R[x])\leq 1+\gldim(R)$. This is achieved by the following exact sequence. Denote $U$ forgetful functor from $R[x]\text{-Mod}$ to $R\text{-Mod}$. Let $M\in R[x]\text{-Mod}$.
$$0\to R[x]\otimes_R U(M)\xrightarrow{\beta} R[x]\otimes_RU(M)\xrightarrow{\times} M\to 0$$
Last map is obviously given by multiplication. $\beta(t\otimes m)=t(x\otimes m-1\otimes xm)$ for any $t\in R[x],\;m\in U(M)$.
$\textbf{Q:}$ Why is $\beta$ naturally defined here? This definition seems very artificial. In other words, I want a concrete description of $\operatorname{Ker}(\times:R[x]\otimes U(M)\to M)$ but $\beta$ turns out not obvious candidate. It seems the whole purpose of $x$ presenting itself in $\beta$ is to shift degree of polynomial and magically allow induction procedure.
Well, $x\otimes m-1\otimes xm$ is obviously an element of the kernel of $\times$. And intuitively, elements of this form ought to "generate" the whole kernel, since the only difference between $R[x]\otimes_R U(M)$ and $R[x]\otimes_{R[x]} M\cong M$ is that the latter knows about scalar multiplication by $x$ in $M$. That is, if you allow $x$ to move across the $\otimes$ symbol, that turns $R$-bilinearity into $R[x]$-bilinearity and should turn $R[x]\otimes_R U(M)$ into $M$. So it is entirely reasonable to guess that the kernel of $\times$ is generated by elements of the form $x\otimes m-1\otimes xm$, and then $\beta$ is just the obvious map of $R[x]$-modules you can write down which contains all those elements in its image. (Note that $\beta$ is just obtained by taking the $R$-linear map $M\to \ker(\times)$ given by $m\mapsto x\otimes m-1\otimes xm$ and then freely making it into a map of $R[x]$-modules.)