This is mostly a response to the title question and the third question in the body; I have nothing intelligent to say about finite products. The point of view I want to defend here is the following:

Simplicial objects are a natural generalization of coequalizer diagrams to higher category theory.

**The story in ordinary category theory**

Let $C$ be an ordinary category and let $F : J \to C$ be a diagram in $C$. For reasons that will hopefully be clear, let me write $J_0$ for the objects of $J$ and $J_1$ for the morphisms. Recall that to compute colimits in $C$ it suffices to be able to compute coproducts and coequalizers, since whenever these exist $\text{colim}(F)$ can be written as the coequalizer of the diagram

$$\bigsqcup_{j_1 \in J_1} F(\text{source}(j_1)) \rightrightarrows \bigsqcup_{j_0 \in J_0} F(j_0)$$

where the two arrows are, respectively, the identity arrows $F(\text{source}(j_1)) \to F(\text{source}(j_1))$ and the compositions $F(\text{source}(j_1)) \xrightarrow{F(j_1)} F(\text{target}(j_1))$. Thus, informally, "colimits are generated by coproducts and coequalizers."

One way to think about this construction is to think of a coequalizer diagram $E \rightrightarrows V$ as being an internal graph in a category, with $E$ being the object of edges and $V$ being the object of vertices, where the two arrows specify the source resp. the target of the edges. The coequalizer, then, is the object of connected components of this internal graph, since we identify the sources of all of the edges with the targets; in other words, it computes $\pi_0$ of the internal graph.

The above decomposition of an arbitrary colimit into coproducts and then a coequalizer is just saying that all colimits are computed by taking the coproduct of a bunch of things, then making some identifications, and that moreover these identifications can all be computed by a single coequalizer which packages together all of the identifications that need to be done using another coproduct. Before taking the coequalizer, the diagram above is an internal graph that describes all of the identifications that need to be done.

**The story in higher category theory**

Now let $C$ be the $(\infty, 1)$-category of spaces, although $C$ can be replaced by any $(\infty, 1)$-category, and let $F : J \to C$ be a diagram in $C$. For ease of exposition let's take $J$ to be an ordinary category although I believe $J$ can in full generality be another $(\infty, 1)$-category, and in particular can be an $\infty$-groupoid, e.g. a space. We now want to compute the $(\infty, 1)$-colimit, or equivalently the homotopy colimit, of $F$ in a way analogous to the coequalizer recipe above.

However, in the homotopical setting it is not enough to be able to compute coproducts and homotopy coequalizers. The problem is that when we take homotopy coequalizers, we never identify things; instead, we add $1$-cells between things, and in particular if we identify things multiple times we end up adding multiple $1$-cells. Some of these $1$-cells should be the "same $1$-cell" and we need to account for this; that is, we should also glue in extra $2$-cells between $1$-cells. But then we run into the same problem one level up and need to glue in extra $3$-cells between $2$-cells, and so forth.

*Example.* Let $J$ be the one-object category corresponding to a discrete group $G$ and let $F : J \to C$ be the trivial diagram. Then the homotopy coequalizer recipe above, when taken in spaces, takes a point and adds to it a loop for every element of $G$. In other words, we get $BF_{|G|}$, the classifying space of the free group on the underlying set of $G$, whereas the correct homotopy colimit is $BG$. The discrepancy is due to the fact that we have failed to account for the relations among the elements of $G$. If we did this by gluing in appropriate $2$-cells then we get a $2$-complex which now has the correct $\pi_1$ but the wrong higher homotopy. The discrepancy is now due to the fact that we have failed to account for the relations among the relations, and so forth.

(If $F$ is a nontrivial diagram, then the homotopy colimit computes the homotopy quotient $F(\text{pt}) \times_G EG$.)

*Example.* To give at least one example where $C$ isn't spaces, let $C$ be chain complexes of abelian groups. Then the correct homotopy colimit of the diagram $J$ from the previous example is (up to quasi-isomorphism) the complex computing the group cohomology of $G$ with coefficients in $\mathbb{Z}$, whereas the homotopy coequalizer only gives the zeroth and first terms $\mathbb{Z}[G] \to \mathbb{Z}$.

(If $F$ is a nontrivial diagram, then the homotopy colimit computes the group cohomology of $G$ with coefficients in $F(\text{pt})$.)

So let's follow our noses. To compute the homotopy colimit $\text{hocolim}(F)$ we might as well write down the diagram

$$\bigsqcup_{j_1 \in J_1} F(\text{source}(j_1)) \rightrightarrows \bigsqcup_{j_0 \in J_0} F(j_0)$$

to start with and then keep on fixing it. As mentioned above, the problem with this diagram is that in its homotopy coequalizer we've added in redundant $1$-cells without specifying $2$-cells relating them. These come from composition of morphisms in $J$: that is, for all composable morphisms $j_1, j_1'$ there's an identification involving $j_1$ and an identification involving $j_1'$ which makes the identification involving $j_1 \circ j_1'$ redundant. So we need to add in $2$-cells corresponding to all pairs of composable morphisms in $J$. Let me write $J_2$ for this set. Then we should extend the diagram above to a diagram

$$\bigsqcup_{j_2 \in J_2} F(\text{source}(j_2)) \substack{\longrightarrow\\[-1em] \longrightarrow \\[-1em] \longrightarrow} \bigsqcup_{j_1 \in J_1} F(\text{source}(j_1)) \rightrightarrows \bigsqcup_{j_0 \in J_0} F(j_0)$$

where the new three arrows correspond to the three morphisms $j_1, j_1', j_1 \circ j_1'$ you can write down starting from a composable pair $j_2 = (j_1, j_1')$ of morphisms, and $\text{source}(j_2)$ denotes $\text{source}(j_1)$ (my convention for composition here is opposite the usual one; for $j_1 \circ j_1'$ to be defined means that the target of $j_1$ is the source of $j_1'$); these are the three $1$-cells bounding a $2$-cell that we now want to put in to fix the redundancy we created.

But of course now we have redundant $2$-cells, and so forth. Hopefully you've anticipated the end of the story by now: in the end we need to consider the entire nerve of $J$, and to compute the homotopy colimit we compute the geometric realization of the simplicial object built from the nerve by applying $F$. Thus we can generalize "colimits are generated by coproducts and coequalizers" to, at least for the case of category-indexed diagrams in spaces, "homotopy colimits are generated by coproducts and geometric realizations."

(I haven't told you what I mean by the geometric realization of a simplicial object in an arbitrary $(\infty, 1)$-category. The tautological definition is that it is the homotopy colimit. In particular, I don't mean the thing you get by starting from an arbitrary cosimplicial object.)

notpreserve finite products. Indeed, for symmetric simplicial sets, $\Delta^n \times \Delta^m \cong \Delta^{n m + n + m}$. – Zhen Lin Jun 16 '14 at 10:05notignore the ordering!! – Todd Trimble♦ Jun 16 '14 at 12:15