Consider an integral domain $R$, its field of fractions $K$ and an $R-$module $M$ that has at least one linearly independent element, i.e.:
$$ D:=\{ m \in M \mid \forall \ r \in R \backslash\{0\}, \ rm \neq 0\} \neq \emptyset.$$
Then it should be true that $M \otimes_R K \neq 0$, but I don't know how to show it. It seems logical that $d \otimes 1 \neq 0$ for all $d \in D$, but how to prove it? We could find an $R-$bilinear map $ \alpha\colon M \times K \to N$ for some $R-$module $N$ so that $\alpha(d,1) \neq 0$, but I cannot find such an $\alpha$.
Here is a general outline:
The first thing to note is that $$K \otimes_R M \cong K \otimes M/M_{tor},$$
where $M_{tor}$ is the torsion submodule of $M$ (this is not hard to show). In your case we have that $M_{tor} \neq M$, so we can reduce to the case where $M$ is torsion-free.
Then the idea is, as you have said yourself, to construct a bilinear map from $K \times M$ to a nonzero $R$-module. To do this, we want to take a look at formal products of the type $x \cdot m$ with $x \in K$ and $m \in M$. Formally, define on $K \times M$ an equivalence relation by
$$(\frac{a}{b}, m) \sim (\frac{c}{d},n) \iff adm = bcn \in M.$$
To show that this is indeed well-defined and an equivalence relation, you need that $M$ is torsion-free. Once you have this, consider $S := K \times M / \sim$ and we have a scalar multiplication $x \cdot m := (x, m)$ as above. Define addition on it in the canonical way and we obtain that $S$ is a $K$-vector space, in particular $R$-module. Check easily that the canonical map $M \to S$ is injective, hence $S \neq 0$ as $M \neq 0$. Then get a bilinear map $K \times M \to S, (x, m) \mapsto x \cdot m$, which can easily be shown to be surjective, and you are done.