$\mathrm{SO}(3)$ is a wonderful group, but it quickly got bored rotating vectors in space. It wanted to be used to help solve differential equations, and so desperately wanted to act on functions. Then one day, it realized it could act on functions by acting on their input.
For instance, it contains a 90 degree clockwise rotation in the $x$-$y$ plane. It takes $(x,y,z)$ and replaces it with $(y,-x,z)$.
Well, the function $2x+3y+7z$ can also be rotated! We just replace $x$ by $y$, $y$ by $-x$, and leave $z$ alone: $2x+3y+7z$ is rotated into $2y+3(-x)+7z = -3x + 2y + 7z$. If we consider the action on all these linear $Ax+By+Cz$ it becomes clear that $\mathrm{SO}(3)$ acts by 3×3 matrices on the vector space with basis $\{x,y,z\}$.
Yay, $\mathrm{SO}(3)$ can act on functions now, and it acts on those linear functions as 3×3 matrices.
What about quadratics? Well we could have $2x^2 + 3xy + 7y^2$. That same 90 degree rotation takes it to $2(y)(y) + 3(y)(-x) + 7(-x)(-x) = 7x^2 - 3xy + 2y^2$. Yay, now $\mathrm{SO}(3)$ acts as 6×6 matrices on the quadratic polynomials $Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2$ spanned by $\{ x^2, xy, y^2, xz, yz, z^2 \}$.
It turns out though that $\mathrm{SO}(3)$ doesn't swirl these functions around very thoroughly. If you take a function, and rotate its input, all it does is rotate the laplacian. In particular, notice that it takes any harmonic polynomial (one with 0 laplacian) to a harmonic polynomial. For quadratic polynomials, this is particularly easy to describe:
When you use $\mathrm{SO}(3)$ to rotate $Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2$, it leaves the laplacian, $A+C+F$, alone. So it will never rotate $x^2$ into $xy$. In fact, it leaves the 5-dimensional space spanned by $\{ x^2-y^2, y^2-z^2, xy, xz, yz \}$ invariant. If we write things in these coordinates, then we get $\mathrm{SO}(3)$ acting as 5×5 matrices. In fact it acts irreducibly.
What happened to the 6th dimension? Well, it is spanned by a very silly function, the sphere: $x^2+y^2+z^2$. If you rotate the sphere, you get the sphere. On this one dimensional space, $\mathrm{SO}(3)$ is represented by 1×1 matrices. Well, actually, matrix. Every single rotation in $\mathrm{SO}(3)$ acts as the identity matrix $[1]$, since every rotation leaves the sphere alone.
If we use colors, like pink, lime, and periwinkle to describe vectors, then a polynomial like $5x^2-5y^2 + 6x + 3$ is composed of a pink term, $5x^2-5y^2$, where $\mathrm{SO}(3)$ acts in a 5×5 manner, a lime term, $6x$, where $\mathrm{SO}(3)$ acts in a 3×3 manner, and a periwinkle term, $3$, where $\mathrm{SO}(3)$ acts in a 1×1 manner.
A polynomial like $7x^2+5y^2$ is a little trickier to see its colors (it is a good thing $\mathrm{SO}(3)$ is so clever), $7x^2+5y^2 = 4(x^2+y^2+z^2) + 3(x^2-y^2) + 4(y^2-z^2)$. The fist term $4(x^2+y^2+z^2)$ is periwinkle where $\mathrm{SO}(3)$ acts in a 1×1 manner, but the next two terms $3(x^2-y^2) + 4*(y^2-z^2)$ are pink where $\mathrm{SO}(3)$ acts in a 5×5 manner.
It is a remarkable and beautiful fact that the irreducible representations of the symmetric group $S_n$ are in correspondence with the partitions of $\lambda \vdash n$. For example, in the case of $S_3$, the irreducible partitions correspond to all the partitions of 3, namely
$$(3) \quad (1,1,1) \quad (2,1)$$
The full story is too long for this post, but details can be found in Chapter 4 of 'Representation Theory: A first course' by Fulton and Harris.
To see the decomposition of the regular representation into its irreducible components is most easily done via character theory. Let $\chi$ be the character of the left regular representation (that is, let $S_3$ act on $K[S_3]$ on the left); $\chi(\sigma)$ is the number of fixed points of the action of $\sigma$ on $K[S_3]$ (as these contribute to the trace of this action). It is plain to see that the only element in $S_3$ which fixes anything in $K[S_3]$ is the identity element $e$, and moreover, this fixes every element in $K[S_3]$. Thus we have
$$\chi(e) = |S_3|=6 \qquad \chi(\sigma)=0 \quad \forall \ \sigma \in S_3 \backslash \{e\}$$
Let $\chi_\lambda$ be the character of the irreducible representation of $S_3$ corresponding to the partition $\lambda \vdash 3$. It is a fact from character theory that the inner product of characters of a representation $A$ and an irreducible representation $B$, defined by,
$$\langle \chi_A, \chi_B \rangle = \frac{1}{|G|}\left( \sum_{g \in G} \chi_A(g)\chi_B(g) \right)$$
gives the multiplicity of $B$ in $A$. For your question then, we need to compute this inner product with the character $\chi_\lambda$. For this we only need to know one more fact: that $\chi_\lambda(e)$ is the dimension of the corresponding irreducible representation of $S_3$ corresponding to $\lambda$. We can now compute
$$\langle \chi, \chi_\lambda \rangle = \frac{1}{|S_3|}\left(\chi(e)\chi_\lambda(e) \right) = \frac{1}{6}(6\cdot \chi_\lambda(e)) = \chi_\lambda(e)$$
We see that the irreducible representation corresponding to $\lambda$ appears in the decomposition of the regular representation exactly the 'dimension of representation' number of times.
Here are the correspondences in your case:
$$\lambda = (3) \rightarrow V_0, \dim = 1$$
$$\lambda = (1,1,1) \rightarrow V_1, \dim = 1$$
$$\lambda = (2,1) \rightarrow V, \dim = 2$$
Therefore
$$k[S_3] = V_0 \oplus V_1 \oplus V^{\oplus 2}$$
as desired.
If this is new to you, then there are a lot of details to check here, all of which can be found in Fulton Harris. You should know that this story works for any $n$ and we have in general that
$$K[S_n] = \bigoplus_{\lambda \vdash n} V_\lambda^{\oplus \dim V_\lambda}$$
where $V_\lambda$ is the irreducible representation of $S_n$ corresponding to the partition $\lambda$ of $n$.
Best Answer
This is essentially @QiaochuYuan's point. If $V\neq 0$ then pick any non-zero vector $v\in V$. The $\mathbb{C}$-linear span of the vectors $\{gv|g\in G\}$ is a non-zero $G$-invariant subspace of $V$. As the set of vectors $\{gv|g\in G\}$ has cardinality $|G|$, their span will have dimension at most $|G|$.
Note this is only true under the assumption $V\neq 0$. That should be in the question somewhere - if not you should mention it to your instructor.