For each $i$ we have a short exact sequence $$ 0 \to I_i \to O_X \to O_{Z_i} \to 0. $$ Let us think about it as about a resolution of $I_i$. Then it follows that $$ I_1 \otimes^L I_2\otimes^L \dots \otimes^L I_k = (O_X \to O_{Z_1}) \otimes^L (O_X \to O_{Z_2}) \otimes^L \dots \otimes^L (O_X \to O_{Z_k}). $$ The transversality condition ensures that $$ O_{Z_{i_1}} \otimes^L O_{Z_{i_2}}\otimes^L \dots \otimes^L O_{Z_{i_p}} \cong O_{Z_{i_1} \cap Z_{i_2} \cap \dots \cap Z_{i_p}}. $$ Therefore we conclude that $I_1 \otimes^L I_2\otimes^L \dots \otimes^L I_k$ is quasiisomorphic to the complex $$ O_X \to \oplus O_{Z_i} \to \oplus O_{Z_i \cap Z_j} \to \dots \to O_{Z_1 \cap Z_2 \cap \dots Z_k}. $$ It follows that $$ I_1\cdot I_2\cdot \dots \cdot I_k = Im(I_1 \otimes^L I_2\otimes^L \dots \otimes^L I_k \to O_X) $$ is equal to the kernel of the map $O_X \to \oplus O_{Z_i}$ that is to $I_{Z_1 \cup Z_2 \cup \dots \cup Z_k}$.
EDIT. The importance of the transversality condition can be seen at the following simple example. Let $X = A^3$ and $Z_1,Z_2$ be two lines intersecting at a point $P$ (the simplest nontransversal intersection). Then one has $O_{Z_1} \otimes O_{Z_2} = O_P$, $Tor_1(O_{Z_1},O_{Z_2}) = O_P$. Therefore one has a spectral sequence with the first term being $$ \begin{array}{ccccc} O_X & \to & O_{Z_1} \oplus O_{Z_2} & \to & O_P \cr &&&& O_P \end{array} $$ which converges to $I_{Z_1} \otimes^L I_{Z_2}$. The second term then gives exact sequence $$ 0 \to I_{Z_1}\otimes I_{Z_2} \to I_{Z_1 \cup Z_2} \to O_P \to 0, $$ which shows that $I_{Z_1}\cdot I_{Z_2}$ is the ideal of the union of lines with an additional nilpotent at the point of intersection.