$f:A\to B$ a ring homomorphism. Then it induces the map $f^*:Spec(B)\to Spec(A)$. $M$ is a finitely generated $A-$module. Let $M=\langle x_1,\dots,x_n\rangle$ We want to show $$Supp(B\otimes M)={f^*}^{-1}Supp(M))$$
Now suppose $M=A x \cong A / \mathfrak{a}$ is cyclic. Then $M_B=B \otimes_A M \cong B / \mathfrak{a} B=B / \mathfrak{a}^e$. Thus $$Supp\left(M_B\right)=V\left(\operatorname{Ann}\left(B /\mathfrak{a}^e\right)\right)=V\left(\mathfrak{a}^e\right)$$ while $$Supp(M)=V(\operatorname{Ann}(A / \mathfrak{a}))=V(\mathfrak{a})$$ $\mathfrak{a}^e \subseteq \mathfrak{q} \Rightarrow \mathfrak{a} \subseteq \mathfrak{a}^{e c} \subseteq \mathfrak{q}^c=\mathfrak{p}$, while $\mathfrak{a} \subseteq \mathfrak{p}=\mathfrak{q}^c \Rightarrow \mathfrak{a}^{\mathfrak{e}} \subseteq \mathfrak{q}^{c e} \subseteq \mathfrak{q}$, so $\operatorname{Supp}\left(M_B\right)=\left(f^*\right)^{-1}(\operatorname{Supp}(M))$ for $M$ cyclic. Now suppose $M$ is generated over $A$ by some $x_i$, so $M_B$ is generated over $B$ by the $1 \otimes x_i$. Write $\mathfrak{a}_i=\operatorname{Ann}_A\left(x_i\right)$.
$$Supp(M_B)= Supp\left( \sum B(1\otimes x_i) \right) =\bigcup Supp(B(1\otimes x_i))\overset{?}{=}\bigcup Supp(B\otimes_A Ax_i)=\bigcup (f^*)^{-1}(Supp(Ax_i))=(f^*)^{-1}\left(\bigcup Supp(Ax_i) \right)=(f^*)^{-1}(Supp(M))$$
Now we have this wrong at the step marked by question mark since tensor is not left exact. The map $B\otimes_A Ax_i\to B\otimes_A M$ induced by the inclusion $Ax_i\to M$ need not be injective. Now why then support of the submodule of $M_B$ generated by $1\otimes x_i$ can easily be smaller than $Supp(B\otimes_A Ax_i)$