Disclaimer: There is nothing about Young tableau in this answer; I realize after rereading your question that your primary question might regard Young tableau; my apologies if this is useless to you.
As a matter of notation, if Lie algebras $\mathfrak g_1$ and $\mathfrak g_2$ are isomorphic, then let's write $\mathfrak g_1\cong\mathfrak g_2$.
In physics, we are interested in the four Lie algebras you listed because the quantum fields and states of particles transform under representations of groups having those algebras.
The spin state of a particle of spin $s$ transforms under the irreducible spin $s$ representation of $\mathrm{SU}(2)$, and we use it's Lie algebra $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ to analyze its representations.
In a Lorentz-invariant Qauntum field theory with Hilbert space $\mathcal H$, we assume in physics that there is a unitary representation $U$ of the Lorentz group $SO(1,3)$ acting on $\mathcal H$ which represents the Lorentz symmetry. This representation acts on the fields $\phi$ in the theory by conjugation
\begin{align}
\phi(x) \to U(\Lambda) \phi(x) U(\Lambda)^{-1}
\end{align}
If $\phi$ is the quantum field describing a certain kind of particle, then this conjugation action will do something special to the indices of the field via a certain finite-dimensional representation of the Lorentz group. The finite-dimensional, irreducible representations $D^{(s_1, s_2)}$ of the Lorentz group are labeled with two "spins" as you have indicated (see below for how this works). But quantum fields don't necessarily transform under on these these.
Example. Dirac Fermion
The electron is a Dirac Fermion. If we denote the corresponding quantum field by $\Psi^a$, then this means that there is a finite-dimensional representation $\rho_\mathrm{D}$ of the Lorentz group such that
\begin{align}
U(\Lambda) \Psi^a(x) U(\Lambda)^{-1} = \rho_\mathrm{D}(\Lambda)^a_{\phantom ab}\Psi^b(\Lambda^{-1} x)
\end{align}
It turns out that the Lie algebraic Dirac representation can actually be written as a direct sum of the left and right handed Weyl spinor representations representations of the Lorentz algebra
\begin{align}
D^{(\frac{1}{2},0)}\oplus D^{(0,\frac{1}{2})}
\end{align}
In general, the indices (target space) of the field corresponding to each kind of particle in a given Lorentz-invariant field theory with transform under a particular finite-dimensional representation of the Lorentz group.
Lie algebraic gymnastics
As you note
\begin{align}
\mathfrak{su}(2) \cong\mathfrak{so}(3)
\end{align}
so both of these algebras equivalently describe the spin state of every particle. In particular, the state of a spin $s$ particle is described by the spin-$s$ irreducible representation $D^{(s)}$ of these algebras.
Next, note that for any $n$, we have an isomorphism $\mathfrak{sl}(n, \mathbb C)\cong\mathfrak{su}(n)_\mathbb C$
In other words, $\mathfrak{sl}(n,\mathbb C)$ is isomorphic to the complexification of $\mathfrak{su}(n)$. In particular, for $n=2$ we have
\begin{align}
\mathfrak{sl}(2, \mathbb C)\cong\mathfrak{su}(2)_\mathbb C
\end{align}
What the heck do complexifications have anything to do with anything you ask? Well, recall that when we construct the representations of $\mathfrak{su}(2)$, we usually do so by defining the ladder operators
\begin{align}
J_{\pm} = J_1 \pm iJ_2
\end{align}
These ladder operators are not actually elements of $\mathfrak{su}(2)$, they are, in fact, elements of its complexification, which, by the isomorphism above is really just the same as $\mathfrak{sl}(2,\mathbb C)$! So what we're really doing with ladder operators, is we're determining irreducible representations of $\mathfrak{sl}(2,\mathbb C)$, and then we're using the following theorem (see Hall's Lie Groups, Lie Algebras, and Representations proposition 4.6):
Let $\mathfrak g$ be a real Lie algebra, then every finite-dimensional complex representation $\pi$ of $\mathfrak g$ has a unique extension to a complex-linear representation $\pi_\mathbb C$ of $\mathfrak g_\mathbb C$ given by
\begin{align}
\pi_\mathbb C(X+iY) = \pi(X) + i\pi(Y)
\end{align}
Furthermore, $\pi_\mathbb C$ is irreducible if and only if $\pi$ is irreducible.
This theorem shows that in order to make an exhaustive study of finite-dimensional representations of $\mathfrak{su}(2)$, it suffices to study instead the complex-linear representations of $\mathfrak{sl}(2,\mathbb C)$.
Next, note that, $\mathfrak{o}(1,3)$ is not isomorphic to $\mathfrak{sl}(2,\mathbb C)$. In fact, you can kind of see from the representation theory that there are two "copies" of the spin $s$ representation in the representations $(s_+, s_-)$ of $\mathfrak{o}(1,3)$ to which you refer. In fact, there is an isomorphism
\begin{align}
\mathfrak{o}(1,3)_\mathbb C \cong \mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C)
\end{align}
To see this in what we do as physicists, recall that in studying the representations of $\mathfrak{o}(1,3)$, we start with the rotation and boost generators $J_i, K_i$, where $i=1,2,3$, and we form the complex linear combinations
\begin{align}
A_i = \frac{1}{2}(J_i + iK_i), \qquad B_i = \frac{1}{2}(J_i - iK_i)
\end{align}
which now an "A-copy" and a "B-copy" of the $\mathfrak{sl}(2,\mathbb C)$ algebra;
\begin{align}
[A_i, A_j] = i\epsilon_{ijk} A_k, \qquad [B_1, B_j] = i\epsilon_{ijk}B_k, \qquad [A_i, B_j] = 0
\end{align}
Why do we do all of this? Well, the original motivation is that we want to study the representations of $\mathfrak{o}(1,3)$ which are important because the particle states in relativistic quantum field theories, for example, basically transform under these representations. But then we notice that if we pass to the complexification of $\mathfrak{o}(1,3)$ (like with did in going from $\mathfrak{su}(2)$ to $\mathfrak{sl}(2,\mathbb C)$ when studying spin $s$ irreps), then the algebra splits into a direct sum of an algebra with itself, namely $\mathfrak{sl}(2,\mathbb C)$ whose representations we are already quite familiar with from our analysis of spin! This allows us to almost immmediately write down the finite-dimensional, complex-linear irreps of $\mathfrak{o}(1,3)_\mathbb C$, and therefore those of $\mathfrak{o}(1,3)$.
