A groups and representations book states:
$[J_i, K_j] = i \epsilon_{ijk} K_k$ tells us that the $K_i$s tarnsform like a vector under the SO(3) subalgebra of SO(4). As an exercise, show that, for example:
$$e^{i \phi J_3} K_1 e^{i \phi J_3} = cos(\phi) K_1 + sin(\phi) K_3$$
I'm confused by exactly what connection they are making here. A vector wouldn't transform like that, in fact it can't even satisfy the above matrix-vector-matrix multiplication. Also how exactly does this come from the commutation relation?
Start by defining $k_1(\phi)=e^{-i\phi J_3}K_1e^{i\phi J_3}$ and $k_2(\phi)=e^{-i\phi J_3}K_2e^{i\phi J_3}$.
Then $$ \frac{d}{d\phi}k_1=e^{-i\phi J_3}(-iJ_3)K_1e^{i\phi J_3}+e^{-i\phi J_3}K_1(iJ_3)e^{i\phi J_3}=k_2(\phi)$$ and $$ \frac{d}{d\phi}k_2=e^{-i\phi J_3}(-iJ_3)K_2e^{i\phi J_3}+e^{-i\phi J_3}K_2(iJ_3)e^{i\phi J_3}=-k_1(\phi)$$ (notice how a negative sign in one of the exponents is crucial), where we have used the commutation relation.
These two equations give you the solutions $$k_1(\phi)=K_1\cos\phi+K_2\sin\phi$$ and $$k_2(\phi)=K_2\cos\phi-K_1\sin\phi.$$
Now, this is "like" the behaviour of a vector, if you write $(K_1,K_2,K_3)$ and rotate this by $\phi$ around the third axis. It is often the case in mathematics that how something transforms is actually more important than what it is.