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Put $L=L_1\cap L_2$, and let $j:L\hookrightarrow M$ be inclusion. The transverse intersection theorem makes $L$ a smooth submanifold without boundary, withIt is compact because it is a closed subset of either compact input submanifold.

Write ${\nu }_r=TM{|}_{L_r}/T{L}_r$. On $L$, consider the bundle mapIts kernel is $TL$, and it is onto because $T{L}_1+T{L}_2=TM$ at every intersection point. Thus the normal bundle of a transverse intersection has the actual bundle isomorphismTransfer the direct-sum complex structure $J_1\oplus J_2$ through this isomorphism. This supplies a complex structure on the normal bundle itself, not merely after stabilization.

Geometrically, multiply the two classes by forming their external product in $M\times M$ and pulling it back along the diagonal $\Delta :M\to M\times M$. The transversality hypothesis is precisely what makes the inverse image of $L_1\times L_2$ under $\Delta$ the submanifold $L$. The normal bundle of this inverse image is the direct sum just calculated. Multiplication of the corresponding Thom classes agrees with the Thom class of this direct sum, so the complex cobordism product represented by transverse intersections isIf the inputs have complex normal ranks $a,b$, the product has degree $2(a+b)$. If they are disjoint, the representative is empty and the product is zero.