$\operatorname{Ann}(M\otimes_A N)=\operatorname{Ann}M+\operatorname{Ann}N$?

This is an excellent question, and your intuition is not too far off.

Note that the counterexamples given in the other answers involve a non-finitely generated module ($\mathbb Q$ thought of as a $\mathbb Z$-module), and constructions like annihilators behave much better for finitely generated modules than for non-finitely generated ones (as a general rule).

E.g. Suppose that $M = R/I$ and $N = R/J$ are cyclic. Then $M\otimes N = R/(I+J),$ and so your conjectured formula is true in that case.

As you observe, always $\operatorname{Ann} M + \operatorname{Ann} N \subset \operatorname{Ann} M\otimes N$. Regarding the converse, what is true for general finitely generated $R$-modules $M$ and $N$ is that $\operatorname{Ann}M\otimes N$ is contained in the radical of $\operatorname{Ann} M + \operatorname{Ann} N$. So your conjecture becomes correct if you restrict to f.g. modules, and take radicals of both sides.

[To see this, use the fact that for a finitely generated module, the radical of $\operatorname{Ann} M$ is equal to the intersection of those prime ideals $\mathfrak p$ for which $\kappa(\mathfrak p) \otimes_R M \neq 0$, where $\kappa(\mathfrak p)$ denotes the fraction field of $R/\mathfrak p$.]

I don't think that we can do much better than this though, in the finitely generated but non-cyclic case.

To see the kind of things that can happen, choose three ideals $I_1,I_2,J,$ and let $M = R/I_1 \oplus R/I_2$ and $N = R/J$. Then $\operatorname{Ann} M = I_1\cap I_2$, $\operatorname{Ann} N = J$, and $M\otimes N = R/I_1+J \oplus R/I_2 + J$, so that $\operatorname{Ann} M\otimes N = (I_1 + J) \cap (I_2 +J).$

Now we always have $(I_1\cap I_2) + J \subset (I_1 + J) \cap (I_2 + J) \subset \operatorname{rad}\bigl((I_1 \cap I_2) + J\bigr),$ but the first inclusion is typically not an equality (because the lattice of ideals in $R$ is always modular, but typically not distributive).

So if you take a counterexample to the distributive property for some lattice of ideals, feeding it into the above example will give a counterexample to the precise form of your conjecture (i.e. equality on the nose, rather than up to taking radicals) involving only finitely generated modules.

The containment $\mathrm{Ann}_A(M)+\mathrm{Ann}_A(N)\subseteq\mathrm{Ann}_A(M\otimes_AN)$ can be proper. For an example, take $A=\mathbf{Z}$, $M=\mathbf{Z}/2\mathbf{Z}$, and $N=\mathbf{Q}$. Then $\mathrm{Ann}_A(M)=2\mathbf{Z}$, and $\mathrm{Ann}_A(N)=0$, but $M\otimes_AN=\mathbf{Z}/2\mathbf{Z}\otimes_{\mathbf{Z}}\mathbf{Q}=0$ because the left tensor factor is killed by $2$ but $\mathbf{Q}$ is divisible. So $\mathrm{Ann}_A(M\otimes_AN)=\mathbf{Z}$.

Consider the $\mathbb{Z}$-modules $M:=\mathbb{Z}/m\mathbb{Z}$ and $N:=\mathbb{Q}$. Then if I am not terribly mistaking:

$\text{ann}M=m\mathbb{Z}$,

$\text{ann}N=0$,

$M\otimes_\mathbb{Z} N=0$.