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Say I have two fundamental groups of nice spaces. There is a natrual map from the free product to the direct product of these groups.

$$\varphi:\pi_1(X)*\pi_1(Y)\to \pi_1(X)\times\pi_1(Y)$$

This can geometrically be thought of as $\pi_1(X\lor Y)\to\pi_1(X\times Y)$. The kernel of this homomorphism is $\ker\varphi=[\pi_1(X),\pi_1(Y)]$.

Is there any geometric way of thinking about this commutator? I am hoping it is some $\pi_1$ of some construction of the spaces $X$ and $Y$.

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    The kernel is a free group, because it acts freely on the Bass-Serre tree of the free product. In particular, it's the $\pi_1$ of the quotient of the Bass-Serre tree by this group, which is a graph; I'm not sure what it's like anyway. (It should be infinitely generated except in trivial cases: $\pi_1(X)$ or $\pi_1(Y)$ trivial, or both finite). – 2017-02-15

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Let $G=\pi_1(X,x_0)$ and $H=\pi_1(Y,y_0)$. Since $[\pi_1(X),\pi_1(Y)]$ is a subgroup of $\pi_1(X) * \pi_1(Y)$, it is the fundamental group of a certain cover $Z$ of $X\lor Y$, with the quotient $\pi_1(X)\times \pi_1(Y)$ acting on $Z$ by Deck transformations.

We can describe $Z$ as follows. Let $p\colon \bigl(\tilde{X},\tilde{x_0}\bigr)\to (X,x_0)$ and $q\colon\bigl(\tilde{Y},\tilde{y_0}\bigr)\to(Y,y_0)$ be universal covers for $X$ and $Y$, respectively. Then $Z$ is the following subspace of $\tilde{X}\times\tilde{Y}$: $$ \bigl(\tilde{X}\times q^{-1}(y_0)\bigr) \cup \bigl(p^{-1}(x_0) \times \tilde{Y}\bigr). $$ Note that $\pi_1(X)\times \pi_1(Y)$ acts on $\tilde{X}\times\tilde{Y}$ componentwise by Deck transformations, and $Z$ is invariant under this action. It is not hard to see that the quotient of $Z$ by this action is homeomorphic to $X\lor Y$, which proves that it is the desired cover.

For example, if $X$ and $Y$ are circles, then $\tilde{X}=\tilde{Y}=\mathbb{R}$ and $p^{-1}(x_0) = q^{-1}(y_0) = \mathbb{Z}$, so $$ Z \,=\, (\mathbb{R}\times \mathbb{Z}) \cup (\mathbb{Z} \times \mathbb{R}). $$ That is, $Z$ is the standard "grid" in the plane. Note that $Z$ covers $S^1\lor S^1$ in an obvious way, with horizontal lines mapping to the first circle and vertical lines mapping to the second circle.

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    Dang, this was really nice Jim – 2017-02-09
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This is what comes to my mind, but it doesn't address your question exactly. If $\pi_2 X = \pi_2 Y = 0$, then you can take $F$ to be the homotopy fiber of the map $X \vee Y \to X \times Y$. Then you will have a short exact sequence $$0 \to \pi_1 F \to \pi_1 (X \vee Y) \to \pi_1(X \times Y).$$ In general, you will have $$\pi_2(X \times Y) \to \pi_1 F \to \pi_1 (X \vee Y) \to \pi_1(X \times Y).$$ The idea is that "short exact sequences" in spaces are usually taken to "long exact sequences" in algebra.