First: for $k\otimes_A M$ to make sense, you have to have an action of $A$ on $k$; that is, $k$ should be an $A$-module in some way. (It could be the trivial $A$ module, $ar=0$ for all $a\in A$ and $r\in k$, though that would make $k\otimes_AM$ the trivial module).
But: in general, if $S$ and $A$ are commutative rings, and you have an action of $A$ on $S$, then for any $A$-module $M$ you get an $S$-module by taking $S\otimes_A M$: this is called extension of scalars (or extension of the base). The action of $s$ on $S\otimes_A M$ is precisely the one you give: given any $s\in S$ and any generator $t\otimes m$, you define $s(t\otimes m) = (st)\otimes m$ and extend linearly.
Explicitly, we have a $A$-multilinear map
$$f\colon S\times S\times M \to S\otimes_A M$$
given by $(s,t,m)\mapsto st\otimes m$. This map is $A$-multilinear:
f(s+s',t,m)&=(s+s')t\otimes m = (st+s't)\otimes m = st\otimes m+s't\otimes m\\
&= f(s,t,m) + f(s',t,m).\\
f(s,t+t',m) &= s(t+t')\otimes m = (st+st')\otimes m = st\otimes m + st'\otimes m\\
&= f(s,t,m) + f(s,t',m).\\
f(s,t,m+m') &= st\otimes(m+m') = st\otimes m + st\otimes m' = f(s,t,m)+f(s,t,m').\\
f(as,t,m) &= (as)t\otimes m = s(at)\otimes m = f(s,at,m)\\
&= a(st)\otimes m = st\otimes am = f(s,t,am)\\
&= a(st\otimes m) = af(s,t,m).
Therefore, the universal property of the tensor product (the definition) says that the map $f$ induces a unique map $S\otimes_A(S\otimes_A M)\to S\otimes_A M$. This map makes $S\otimes_A M$ into an $S$-module.
In the particular case where $S$ is a field, as you have, this makes $k\otimes_A M$ into a $k$-module; and $k$-modules are the same thing as $k$-vector spaces.