Under which transformation of the Lorentz group do 4-momenta transform?

Any $4$-vector forms an irreducible representation of the Lorentz group, since any Lorentz transformation mixes all four components. But from the point of the $SO(3)$ subgroup it is reducible since spatial rotations do not mix the temporal component $V^0$ with the spatial components $V^i$ of the $4$-vector.

Clearly the temporal component is invariant under a spatial rotation and so it's a singlet (which we indicate by its dimension $mathbf{1}$) under $SO(3)$, while the spatial components form an irreducible three-dimensional rep of $SO(3)$ so they form a triplet $mathbf{3}$. This gives the decomposition under $SO(3)$ that you give in your question $$mathbf{4}=mathbf{1}oplusmathbf{3}$$ which one can write also as $mathbf{0}oplusmathbf{1}$, it's just another convention.

What we've seen up to here is the representation of a $4$-vector under a subgroup of the Lorentz group.

If we want to know in which irrep of the whole Lorentz group a vector belongs. For this we use the following isomorphism between the complexified algebras $$mathfrak{so}(3,1)_mathbb{C}congmathfrak{su}(2)_mathbb{C},oplus,mathfrak{su}(2)_mathbb{C}$$ which physicists, somewhat wrongly, see as an isomorphism between groups. Using this isomorphism we can classify irreps of the Lorentz group using "two" irreps of the $SU(2)$ group. For this we use the convention $(m,n)$. $4$-vectors belong to the $left(dfrac{1}{2},dfrac{1}{2}right)$ rep.

In fact in $SU(2)$ we have the following decomposition $$mathbf{2}otimesmathbf{2} = mathbf{1}oplusmathbf{3}$$ where the $mathbf{2}$ is the fundamental rep of $SU(2)$ and so the rep of spin half. The relation between elements of the $left(dfrac{1}{2},dfrac{1}{2}right)$ and complex vectors is easy to see.

Take the following generic element of the $left(dfrac{1}{2},dfrac{1}{2}right)$ rep $$left((psi_L)_alpha, (xi_R)_betaright)$$ where $psi_L,xi_R$ are two independent Weyl spinors and $alpha,beta$ are spinor indices that take values $1,2$. Now you can easily see that this has four independent components, just like a $4$-vector. Moreover, if we define the matrices $$sigma^mu = (1,sigma^i)qquadbar{sigma}^mu = (1,-sigma^i)$$ where $sigma^i$ are the Pauli matrices and $1$ is the $2times 2$ identity matrix, then $$xi_R^daggersigma^mupsi_Rqquad xi_L^daggerbar{sigma}^mupsi_L$$ are contravariant four vectors. Here we've used the fact that $$xi_L = -isigma^2xi_R^*qquad psi_R = isigma^2psi_L^*$$ These four vectors are, by construction, complex. Since we want them to be real, we have to impose the reality condition $V_mu = V_mu^*$.

Therefore we have obtained the real four-vector representation.

TL;DR: Four vectors transform both as a $mathbf{0}oplusmathbf{1}$ rep of the $SO(3)$ (spatial rotations) subgroup of the Lorentz group and as the $(frac{1}{2},frac{1}{2})$ rep of the $SU(2)times SU(2)cong SO(1,3)$ group.