Creo que esto funciona. Deje $A=k[x,y,z]/(x^2,xy,xz)$, vamos a $M$ ser el cociente de la libre $A$-módulo generado por $e_1$ $e_2$ sujeto a las relaciones $$xe_1=ye_2 \qquad ze_2=0$$ and let $N$ be the quotient of the free $A$-module generated by $f_1$ and $f_2$ subject to the relation $$yf_1=zf_2.$$ Then $e_1$ is free in $M$ and $f_1$ is free in $N$, yet $$x\cdot e_1\otimes f_1 = xe_1\otimes f_1 = ye_2\otimes f_1 = e_2\otimes yf_1 = e_2\otimes zf_2 = ze_2\otimes f_2 = 0.$$
Let me check using Macaulay2:
First, construct our base ring
i1 : R = QQ[x,y,z]/(x*z,x*y,x*x);
Next, $M$ as a quotient of the free module $F=R^2$
i2 : F = R^2;
i3 : M = F / (x*F_0 - y*F_1, z*F_1);
and then $N$, also as a quotient of $F$,
i4 : N = F / (y*F_0 - z*F_1);
The element $e_1$, the image of the first generator of $F$ in $N$ is free:
i5 : kernel map(M, R^1, {{1}, {0}})
o5 = image 0
Likewise, $f_1$,the image of the first generator of $F$ in $N$ is free:
1
o5 : R-module, submodule of R
i6 : kernel map(N, R^1, {{1}, {0}})
o6 = image 0
1
o6 : R-module, submodule of R
Finally, $e_1\otimes f_1$ is not free in $M\otimes N$:
i7 : kernel map(M**N, R^1, {{1}, {0}, {0}, {0}})
o7 = image | x |
1
o7 : R-module, submodule of R
This not only shows that $x$ kills $e_1\otimes f_1$ but that in fact it generates its (one-dimensional) annihilator.
N.B. I constructed this by first deciding the relations which define the modules, and then iteratively computing kernels using and adding relations to the ring until I got $e_1$ and $f_1$ a ser libre.