In general, if $G$ acts on some complex vector space $W$, and you want the $\chi$-isotypical component of $W$, where $\chi$ is some irreducible character, you use the idempotent corresponding to $\chi$: define
$$
e_{\chi} = \frac{\chi(1)}{|G|}\sum_{g\in G}\chi(g^{-1})\cdot g.
$$
This is an element of the group algebra $\mathbb{C}[G]$; in fact it is a primitive central idempotent in $\mathbb{C}[G]$.
Now, $W$ is a module under this group algebra, so you can consider the subspace of $W$ given by $e_{\chi}W = \{e_{\chi}(w): w \in W\}$. This is precisely the $\chi$-isotypical component of $W$. Of course, this is the same as the space spanned by $e_{\chi}(w_i)$ as $w_i$ runs over a basis of $W$.
In your example, $G$ is $S_3$ and $W$ is $V^{\otimes 3}$, and $G$ acts on elementary tensors by permuting the entries. If you write out the above in the case that $\chi$ is the trivial character, you get exactly what you wrote down (strictly speaking, my version differs from yours by the normalisation $1/|S_3|$, which doesn't matter when it comes to linear spans). For the sign character, this procedure gives you
$$
e_{\chi}(v_1\otimes v_2\otimes v_3) = \frac{1}{|G|}\sum_{\sigma\in S_3}{\rm sign}\sigma\cdot v_{\sigma^{-1}(1)}\otimes v_{\sigma^{-1}(2)}\otimes v_{\sigma^{-1}(3)}.
$$
Just like in the case of the trivial character, if you do this for each basis element of $V^{\otimes 3}$, you get redundancies, so to get a basis, you put suitable restrictions on your indices.
See if you can work out the rest yourself; if not, then feel free to ask for more details. It is also a nice exercise to then go on to compute the characters of these components in terms of the character of $V$, like you do in the $V^{\otimes 2}$ case.
You're entirely right in noticing this connection, and this is one place (of many) where some category theory makes the reason for the connection obvious.
We can represent a group $G$ as a category with one object (see here, for instance), call it $BG$. When we do this, a group action $G \curvearrowright X$ is exactly a functor $\alpha : BG \to \mathsf{Set}$ with the unique object in $BG$ sent to $X$.
More generally, if $X$ is an object of a category $\mathcal{C}$, then an action of $G$ on $X$ is a functor $\alpha : BG \to \mathcal{C}$ where the unique object is sent to $X$. For example, this gives continuous actions or smooth actions, by taking $\mathcal{C}$ to be the category of topological or smooth spaces, respectively, or linear actions (which we usually call representations) by taking $\mathcal{C}$ to be a category of modules or vector spaces.
So now say we have an action of $G$ on a set $X$, that is, we have a functor $\alpha : BG \to \mathsf{Set}$. Recall there are also functors $\mathsf{Set} \to k\text{Vect}$ sending a set $X$ to a certain $k$ vector space!
- We can send a set $X$ to the vector space $k^X$ of functions $X \to k$, with pointwise operations. Notice this is contravariant (do you see why?)
- We can send a set $X$ to the vector space $kX$ whose basis is given by elements of $X$.
Notice this is covariant (again, do you see why?)
But now, since a group action (or a representation) is a functor out of $BG$, we can turn our action of $G$ on $X$ into an action of $G$ on $k^X$ (or $kX$): Compose the functors!
$$
BG^\text{op} \overset{\alpha^\text{op}}{\longrightarrow} \mathsf{Set}^\text{op} \overset{k^{-}}{\longrightarrow} k\text{Vect}
$$
$$
BG \overset{\alpha}{\longrightarrow} \mathsf{Set} \overset{k[-]}{\longrightarrow} k\text{Vect}
$$
Now, you're considering the first case. Notice, because $k^{-}$ is a contravariant functor that we had to take the opposite of our action $\alpha$ from $BG$ to $\mathsf{Set}$. As a quick exercise, you should convince yourself that this "opposite"-ness is the reason
$(gf)(x) = f(g^{-1}x)$.
We could also consider the second case, and for completeness, I'll say something about it.
The idea here is that each $x \in X$ is a basis element of $kX$. So each function $g \cdot - : X \to X$ extends linearly to a map on $kX$. Notice when we take this definition, because $k[-]$ is covariant, we don't have to worry about inverting $g$:
$$
g(x_1 + x_2) = gx_1 + gx_2
$$
As for why we call both of these things "representations", as I understand it, we historically thought of groups $G$ defined axiomatically as "abstract groups", and we used the word "representation" to indicate that we were representing an abstract group by something "concrete", like a group of symmetries of some object, a group of matrices, etc. This is why
the theorem that each $G$ embeds into a symmetric group is sometimes called
Cayley's Representation Theorem. Nowadays, though, we usually think of linear representations (that is, actions of $G$ on some module or vector space) when we think of representation theory, and so I believe this older terminology is falling out of fashion.
I hope this helps ^_^
Best Answer
The trivial representation of $S_2$ doesn't have to act on $P$.
Let's say we are looking at the permutation representation $\pi : S_3 \to \operatorname{Aut}(\Bbb{C}^3)$ acting on canonical vectors as $\sigma(e_i) = e_{\sigma^{-1}(i)}$ for $1 \le i \le 3$ and $\sigma \in S_3$.
Now look at the trivial representation $S_2 \to \operatorname{Aut}(\Bbb{C})$ and let's construct the induced representation. We have to select a complete set of representants $\sigma_1, \sigma_2, \sigma_3 \in S_3$ of $S_3/S_2$. We can take $$\sigma_1 = (1 \ 3), \quad \sigma_2 = (2 \ 3), \quad \sigma_3 = \operatorname{id}$$ where $\sigma_i$ represents the coset of maps $\sigma \in S_3$ such that $\sigma(3) = i$.
The induced representation acts on the space $\sigma_1\Bbb{C} \oplus \sigma_2\Bbb{C} \oplus \sigma_3\Bbb{C} \cong \mathbb{C}^3$ such that for $\sigma \in S_3$ and $1 \le i \le 3$ we find $1 \le j(i) \le 3$ and $\tau_i \in S_2$ such that $$\sigma\sigma_i = \sigma_{j(i)}\tau_i$$ and then define $$\sigma(z_1,z_2,z_3) = \sigma(\sigma_1z_1+\sigma_2z_2+\sigma_3z_3) = \sigma_{j(1)}(\tau_1z_1) + \sigma_{j(2)}(\tau_2z_2) + \sigma_{j(3)}(\tau_3z_3).$$
In our case we have $$(\sigma\sigma_i)(3) = \sigma(\sigma_i(3)) = \sigma(i)$$ which implies that $\sigma\sigma_i$ is in the coset represented by $\sigma_{\sigma(i)}$, or $\sigma\sigma_i = \sigma_{\sigma(i)} \tau_i$ for some $\tau_i \in S_2$. Since $S_2$ acts trivially, we have
$$\sigma(z_1,z_2,z_3) = \sigma_{\sigma(1)}(z_1) + \sigma_{\sigma(2)}(z_2) + \sigma_{\sigma(3)}(z_3) =(z_{\sigma^{-1}(1)}, z_{\sigma^{-1}(2)}, z_{\sigma^{-1}(3)})$$ so it acts as $\sigma(e_i) = e_{\sigma(i)}$, i.e. it is precisely the permutation representation $\pi$.