Homotopy groups of a wedge sum

Let $X_\alpha$ be connected CW-complexes. Then from Hatcher's book,

$$\pi_{n}(\prod_{\alpha} X_{\alpha})=\prod_{\alpha}\pi_{n}(X_{\alpha}).$$

Is it true in general that

$$\pi_{n}(\bigvee_{\alpha} X_{\alpha})=\prod_{\alpha}\pi_{n}(X_{\alpha})?$$

No. In fact, the homotopy groups of the wedge sum are often vastly more complicated.

For example, consider $S^1\vee S^2$. Its universal cover looks like a line with an $S^2$ wedged at every integer point. In particular, $\pi_2$ of the universal cover is infinitely generated (one map for each of the countably many spheres), despite the fact that $\pi_2(S^1) = 0$ and $\pi_2(S^2) = \mathbb{Z}$.

Another example may be found in Bott and Tu's book "Differential forms in algebraic topology" near page 264.

There, they compute some of the homotopy groups of $S^2\vee S^2$. In particular, $\pi_4$, $\pi_5$, and $\pi_6$ are are infinite, even after tensoring with $\mathbb{Q}$. But for $S^2$, only $\pi_2$ and $\pi_3$ are infinite.

No. In general, homotopy groups behave nicely under homotopy pull-backs (e.g., fibrations and products), but not homotopy push-outs (e.g., cofibrations and wedges). Homology is the opposite.

For a specific example, consider the case of the fundamental group. The Seifert-Van Kampen theorem implies that $\pi_1(A\vee B)$ is isomorphic to the free product $\pi_1(A)*\pi_1(B)$, not the product.

No, this follows for example from Hilton's Theorem