Let $R$ be an arbitrary ring, with $M$ and $N$ being $R$-modules.
If I have an element:
$$x := \Sigma_{i=1}^k r_i \mu_i \otimes n_i \in M^* \otimes N$$
such that:
$$\Sigma_{i=1}^k r_i \mu_i(m) \space n_i \in N$$
is zero for all $m \in M$, does this imply that $x = 0$?
No, not necessarily.
As an example, take $R=\mathbb{Z}/4\mathbb{Z}$ and consider the $R$-modules $M=N:=2R$. Then $M^*\simeq M$ and thus $M^*\otimes_R N\simeq M$ again (which is all straightforward to check directly). In particular, the element $x=\iota\otimes 2$ (where $\iota:M\hookrightarrow R$ is the inclusion) is non zero. But for any $m\in M$ we nonetheless have $$\iota(m)\times 2=2m\in 4R=0,$$ so that all evalutations vanish.
Edit: Here is an example with $R$ an integral domain: Let $R=\mathbb{Z}[X]$ and consider the ideal $J=(2,X)$. We let $M=J$ and $N=R/J$. I first claim that $J^*\simeq R$ and that every functional is of the form $$\phi_r:J\to R, j\mapsto rj.$$ Indeed, suppose that $\phi:J\to R$ is $R$-linear. Then we must have $$\phi(2X)=2\phi(X)=X\phi(2).$$ Since both 2 and $X$ are prime in $R$, we must have $\phi(X)=Xr$ and $\phi(2)=2s$ for some $s,r\in R$. But then immediately $2Xr=2Xs$, so that $r=s$ and $\phi(j)=rj$ for each $j$ by linearity.
In particular we have $$J^*\otimes N\simeq R\otimes N=N,$$ which is non zero. But from the above, the image of every functional is contained in $J$, from which it follows that every tensor evaluates to zero, since $$\phi\otimes n(j)=\phi(j)n\in JN=0.$$
Hence, this phenomenon is not tied to zero divisors. I do suspect that the answer might change for $R$ a Dedekind domain (since this type of argument falls apart when every ideal is invertible), but I have not checked this.