From Rotman's Algebraic Topology concerning the wedge operator:
If $K_1$ and $K_2$ are simplicial complexes, then for all $n \ge 1$, $$H_n(K_1 \lor K_2) \cong H_n(K_1) \oplus H_n(K_2)$$
What is the definition of the wedge of two simplicial complexes $K_1 \lor K_2$? Directly above this problem Rotman defines the wedge of two pointed spaces to be the quotient space of their disjoint union with identified basepoints.
If we take $K_i$ as pointed spaces then they're a finite disjoint union of elements $s$ (simplexes of $K_i$) and therefore their nontrivial homology groups should be $0$, but it may be the case that $H_n(K_i)$ is not $0$ so I do not see how this could be an appropriate definition.
A wedge of pointed simplicial complexes is obtained (as for topological spaces) by identifying the two "points," which in this case are 0-simplices. With this definition, $|K_1| \vee |K_2|$ is homeomorphic to $|K_1 \vee K_2|$.
By 'identifying the two "points," ' I mean: we may assume that $K_1$ and $K_2$ are disjoint. If $v_1$ and $v_2$ are the distinguished 0-simplices in $K_1$ and $K_2$, then you create a new simplicial complex, replacing both of these with a single 0-simplex $w$, and the simplices in this new complex are the simplices in either $K_1$ or $K_2$, except you replace $v_i$ with $w$ wherever it appears. (I'm viewing simplices as being defined by their vertices, so if $(v_1, x, y, z)$ is a simplex in $K_1$, then $(w, x, y, z)$ will be a simplex in $K_1 \vee K_2$.)