You may have been confused by the fact that you have a preconceived idea of how $S_3$ should act on the components. Keep in mind that there is no natural action of $S_N$
on any of the modules $V(\lambda)$. That action comes to being only when we look at subspaces of $V^{\otimes N}$ specifically! So Schur-Weyl seeks to identify subspaces of $V^{\otimes N}$, and identify what they look like as reps for $\mathfrak{sl}_3$ and $S_N$ independently from each other.
I will try and describe this decomposition in the case $N=3$. I use weight diagrams. Here's the way I draw $V$:
The black dots are the weights. The red arrows are the roots. The blue arrows are the fundamental dominant weights. The green circles give the formal character of the fundamental $\mathfrak{sl}_3$-module $V=V(\lambda_1)$. A single green circle means that the multiplicities of all the weights are $=1$.
The second diagram shows the formal character of $V^{\otimes 3}$. Notice that this time multiplicities $3$ and $6$ occur, and I try to convey that with the appropriate number of concentric green circles.
We immediately spot the highest weight $3\lambda_1$. Indeed, the 10-dimensional $\mathfrak{sl}_3$-module $V(3\lambda_1)$ is a summand of $V^{\otimes3}$.
This summand consists of totally symmetric tensors, so it is trivial as an $S_3$-module. What the Schur-Weyl formula is trying to convey is that the 10-dimensional subspace $W_1$ has the structure $V(3\lambda_1)\otimes \mathbf{1}$ when viewed as a mixed module of $\mathfrak{sl}_3\times S_3$. Recall that the formal character of $V(3\lambda_1)$ looks like
In other words, all its weights have multiplicity one.
Next we observe that the second highest weight of $V^{\otimes 3}$ is $\lambda_1+\lambda_2$, appearing with multiplicity $3$. We recall that this is the highest weight of the adjoint representation of $\mathfrak{sl}_3$, of dimension $8$.
The conclusion is that the adjoint representation appears as a composition factor of $V^{\otimes3}$ with multiplicity two. So there is a $16$-dimensional subspace $W_2$ of
$V^{\otimes3}$ that as an $\mathfrak{sl}_3$ module looks like two copies of $V(\lambda_1+\lambda_2)$. What does it look like as an $S_3$-module? If the weight vectors of $V$ are $x_1$ of weight $\lambda_1$,$x_2$ of weight $-\lambda_1+\lambda_2$, and $x_3$ of weight $-\lambda_2$, then the weight space corresponding to weight $\lambda_1+\lambda_2$ in $V^{\otimes3}$ is the span of
$$x_1\otimes x_1\otimes x_2, x_1\otimes x_2\otimes x_1, x_2\otimes x_1\otimes x_1.$$
We see that the group $S_3$ permutes those three vectors according to its natural $3$-dimensional representation. We recall from representation theory of finite groups that this $3$-dimensional rep splits into a direct sum of the trivial representation
(here spanned by the averagre of those three vectors belonging to the subspace $W_1$ of symmetric tensors), and a 2-dimensional irreducible representation, call it $M$. Obviously the space $W_2$ must then be
a bunch of copies of $M$ as an $S_3$-module. In other words, Schur-Weyl wants us to identify the space $W_2$ as
$$W_2=V(\lambda_1+\lambda_2)\otimes M.$$
As a refresher please find the weight diagram of the adjoint representation:
Finally, there is the 1-dimensional subspace $W_3$ of completely antisymmetric tensors. Because the weights of $V$ add up to zero, $W_3\cong V(0)$ as an $\mathfrak{sl}_3$-module. Because the tensors are totally antisymmetric, as a representation of $S_3$ we see that $W_3$ looks like $\mathbf{alt}$, the $1$-dimensional representation affording the sign character.
So the Schur-Weyl decomposition looks like
$$
\begin{aligned}
V^{\otimes3}&=\left(V(3\lambda_1)\otimes \mathbf{1}\right)\\
&\oplus \left(V(\lambda_1+\lambda_2)\otimes M\right)\\
&\oplus \left(V(0)\otimes\mathbf{alt}\right).
\end{aligned}
$$
As a check let's verify differently that $V^{\otimes3}$ splits in the prescribed way as an $S_3$-module. Let $\psi$ be the character of $V^{\otimes3}$ as an $S_3$-module. We see that
- $\psi(1_{S_3})=27$.
- The permutation $(12)$ fixes the nine basic tensors of the form $x_i\otimes x_i\otimes x_j$, $i=1,2,3$, $j=1,2,3$, and permutes the other vectors of the basis. Meaning that $\psi((12))=9$ and the same holds for all the other 2-cycles.
- The 3-cycle $(123)$ fixes the three basis vectors $x_i\otimes x_i\otimes x_i$ so $\psi((123))=3$.
- The multiplicity of the trivial character $\chi_0$ as a component of $\psi$ is thus
$$\langle\psi,\chi_0\rangle_{S_3}=\frac16(1\cdot27\cdot1+3\cdot9\cdot1+2\cdot3\cdot1)=10.$$
- Similarly the multiplicity of the $2$-dimensional character $\chi_M:1\mapsto 2, (12)\mapsto0,(123)\mapsto-1$ is
$$\langle\psi,\chi_M\rangle_{S_3}=\frac16(1\cdot27\cdot2+3\cdot9\cdot0+2\cdot3\cdot(-1))=8.$$
- And the multiplicity of the sign character $\mathbf{alt}$ is
$$\langle\psi,\mathbf{alt}\rangle_{S_3}=\frac16(27\cdot1+3\cdot9\cdot(-1)+2\cdot3\cdot1)=1.$$
All of this matches with the conclusion that $W_1$, $W_2$ and $W_3$ are also the isotypic components of $V^{\otimes 3}$.
