Let $f:R \to S$ be a homomorphism of rings, and let $M$ be a module over $R$. Consider the tensor product $M \otimes_R S$, where $S$ is regarded a left $R$-module via $f$. Since ${\displaystyle S}$ is also a right module over itself, and the two actions commute, that is ${\displaystyle r\cdot (s\cdot s')=(r\cdot s)\cdot s'}$ for ${\displaystyle r\in R},$ $ {\displaystyle s,s'\in S}$. ${\displaystyle M \otimes_R S}$ inherits a right action of $ {\displaystyle S}$. It is given by ${\displaystyle (m\otimes s)\cdot s'=m\otimes ss'}$ for $ {\displaystyle m\in M}$, ${\displaystyle s,s'\in S}$. This module is said to be obtained ${\displaystyle M}$ through extension of scalars.
Could someone elaborate for me what is the role of $f$ here? I presume that this is supposed to be a tensor product of modules and I do not entirely understand the statement that $S$ is regarded a left $R$-module via $f$. Does this mean that $rs := f(r)s$? If so, then what does it mean for the elements of $M \otimes_R S$, can I multiply them by $r$ via $$r(m\otimes s)=f(r)(m\otimes s)?$$
Yes, this is exactly what this means. The notation is a bit abusive, especially if $f$ is not injective.
Regarding left multiplication by $r\in R$, I don't know the source of your quote, but I think $M$ is meant to be a right $R$-module and $M\otimes_RS$ is a right $S$-module. If $M$ happens to be an $R$-bimodule, then $M\otimes_R S$ is still a left $R$-module, but there isn't any left $S$-module structure on it (unless the rings are all commuatative).