In general, we have a map $\mu:M^\vee\otimes N^\vee\to (M\otimes N)^\vee$ given by $\mu(\phi\otimes\psi)(\sum_i m_i\otimes n_i)=\sum_i\phi(m_i)\psi(n_i)$; this is presumably what mephisto is referring to, and it is an isomorphism in the case that he mentions. If the ring $A$ is a principal ideal domain and $M$ and $N$ are both finitely generated then the map is again an isomorphism, for rather uninteresting reasons, because the functor $M\mapsto M^\vee$ kills all torsion modules.
Now let $M$ be the free module over $A$ with basis $\{e_n\}_{n\in\mathbb{N}}$ and take $N=M$ and define $\xi:M\otimes M\to A$ by $\xi(e_i\otimes e_i)=1$ and $\xi(e_i\otimes e_j)=0$ for $j\neq i$. I claim that this is not in the image of $\mu$. Indeed, if $\zeta=\mu(\sum_{i=1}^r\phi_i\otimes\psi_i)$ then the matrix $(\zeta(e_i\otimes e_j))_{i,j=0}^k$ has rank at most $r$ for all $k$, and it is clear that $\xi$ does not have this property.
For another interesting example, take $A=k[x,y]/(xy)$ and $M=A/x$ and $N=A/y$. Multiplication by $y$ gives a map $\phi:M\to A$, so $\phi\in M^\vee$. Clearly $x\phi=0$ so multiplication by $\phi$ gives a well-defined map $M=A/x\to M^\vee$, which is an isomorphism. Similarly we have an isomorphism $N\to N^\vee$, so $M^\vee\otimes N^\vee\simeq M\otimes N=A/(x,y)=k$. On the other hand, $(M\otimes N)^\vee=\text{Hom}_A(A/(x,y),A)=0$. Thus, we have a case where $\mu:M^\vee\otimes N^\vee\to(M\otimes N)^\vee$ is not injective.