Question: "Namely, how do derive eq (2) from the given eq (1) for the Lie algebra matrix representations?"
Answer: Let $k$ be the real numbers and $K$ the complex numbers and let $G:=SU(2)$ be the special unitary group. It follows its complexification $G_K\cong SL(2,K)$ is the special linear group. The irreducible representations of $SL(2,K)$ are classified. Let $V:=K\{e_1,e_2\}$ be the standard representation of $SL(2,K)$, with dual $V^*:=K\{x_1,x_2\}$. There is a canonical map of $SL(2,K)$-modules
$$ \wedge: V\otimes_K V \rightarrow \wedge^2 V \cong T$$
and since $T$ is the trivial $SL(2,K)$-module it follows $V \cong V^*$ is an isomorphism of $SL(2,K)$-modules. The irreducible $SL(2,K)$-modules are the symmetric powers
$$Sym^d(V) \cong Sym^d(V^*)$$
for $d \geq 0$. "Using" the isomorphism $G_K \cong SL(2,K)$ it follows the irreducible $SL(2,K)$-modules $Sym^d(V),Sym^d(V^*)$ are irreducible $G$-modules with dimension
given by
$$dim_K(Sym^d(V))=d+1, dim_k(Sym^d(V))=2(d+1).$$
There are precise relations between irreducible $G$-modules and the irreducible $SL(2,K)$-modules (similar for $SU(3)$ and $SL(3,K)$). The irreducible representations of $SL(n,K)$ have been classified using the "Schur-Weyl" construction. Given any "partition" $\lambda$, there is a Schur-Weyl functor $\mathbb{S}_{\lambda}$ and when you apply this functor to the standard representation
$$V(\lambda):=\mathbb{S}_{\lambda}(V)$$
of $SL(n,K)$ you get all finite dimensional irreducible $SL(n,K)$-modules.
You can find some information in
Hall, Brian - Lie Groups, Lie Algebras, and Representations: An Elementary Introduction (Graduate Texts in Mathematics, 222)
and
Fulton/Harris - Representation Theory: A First Course.
Your comment: "For example, for SU(2), we know the tensor product of 2-dimensional fundamental representation of SU(2) gives the 3-dimensional adjoint representation of SU(2):"
Answer: If $dim_k(V)=2$ it follows $dim_k(V^{\otimes_k n}) = 2^n$. You cannot
tensor a 2-dimensional representation with itself and get an odd dimensional representation. The representation $V\otimes V$ may decompose into a direct sum $V_3 \oplus V_1$ with $dim_k(V_i)=i$: The tensor product of two irreducible representations is not necessarily irreducible.
Groups such as $SL(n,K)$ and $SU(n)$ are semi simple Lie groups and they have the property that any finite dimensional $k$ or $K$ representation $V$ decompose into a direct sum
$$V \cong \oplus_i V_i$$
where $V_i$ is an irreducible module. The irreducible modules are classified by their "highest weight vector" and "highest weight". Hence if you want to check if two irreducible modules are isomorphic you must calculate their weights and compare. See the above books for more details.
Example: In your case you have given a basis for $\mathfrak{g}:=\mathfrak{su}(2,k):=k\{\sigma_i\}$
and an explicit representation
$$\rho: \mathfrak{g} \rightarrow End_k(k^2)$$
This induce an explicit adjoint representation
$$ad: \mathfrak{g} \rightarrow End_k(\mathfrak{g})$$
defined by $ad(x)(y):=[x,y]$. Since $dim_k(\mathfrak{g})=3$ you will get explicit matrices $ad(\sigma_i)$ for $i=1,2,3$, and you must calculate these matrices in your given basis $B$. If $x:=\sigma_1$ you must calculate the vectors
$$v_i:=[x, \sigma_i]$$
\begin{align*} [v_i]_B:=\begin{pmatrix} v_{1i} \\ v_{2i} \\ v_{3i} \end{pmatrix} \end{align*}
and construct the matrix
\begin{align*} M_1:= \begin{pmatrix} 0 & v_{12} & v_{13} \\ 0 & v_{22} & v_{23} \\ 0 & v_{32} & v_{33} \end{pmatrix} \end{align*}
Do this for $\sigma_2, \sigma_3$, construct $M_2,M_3$ and compare to the matrices you write down above. You must first check that the basis you give above is a basis for $\mathfrak{su}(2,k)$ as $k$-vector space.
