Please compute through an example before asking! What happens if $U = \text{Spec}(k[x])$ and $Z = V(x)$?

Later edit. The answer to your question is that if $A$ is a ring and $f \in A$ an element, and $M$ an $A$-module, then we can look at the sequence
$$
0 \to M[f^\infty] \to M \to M_f \to M_f/M \to 0 \to 0 \to \ldots
$$
This sequence will be isomorphic, as a sequence of $A$-modules, to the sequence
$$
0 \to H^0_Z(X, \mathcal{F}) \to H^0(X, \mathcal{F}) \to H^0(U, \mathcal{F}) \to
H^1_Z(X, \mathcal{F}) \to H^1(X, \mathcal{F}) \to \ldots
$$
where $X = \text{Spec}(A)$, $Z = V(f)$, $U = X \setminus Z$, and $\mathcal{F}$ is the quasi-coherent $\mathcal{O}_X$-module associated to $M$. Since the functor $\otimes_A A/fA$ is right exact and since $M_f \otimes_A A/fA = 0$ we find that the tensor product you wrote down is zero.

In the above I am using cohomology with supports $H^i_Z$. This can be defined for an arbitrary closed subset $Z$ of a topological space $X$. This then gives me long exact sequences relating $H^i_Z(X, -)$, $H^i(X, -)$, $H^i(U, -)$. If $-$ is an abelian sheaf, then these are long exact sequences of abelian groups. If $-$ is a sheaf of $\mathcal{O}_X$-modules, then these are long exact sequences of $\Gamma(X, \mathcal{O}_X)$-modules. You can prove this by taking a resolution by injectives in the category of $\mathcal{O}_X$-modules and using general theorems saying that such resolutions compute the right thing (i.e., the same as what you get for abelian sheaves). Or you can use the functoriality of the construction (because multiplying by an global section of the structure sheaf commutes with everything in sight).