To prove that $V_1$ is irreducible, it is sufficient to prove that $\mathrm{End}_{\mathbb{C}G}(V_1)$ is one-dimensional. (Subproof: if $V_1$ has a non-zero proper subrepresentation $W$ then $V_1 = W \oplus C$ for some complementary representation $C$. The projection maps onto $W$ and onto $C$ corresponding to this decomposition are linearly independent.)
Let $\theta \in \mathrm{End}_{\mathbb{C}G}(V_1)$. Extend $\theta$ to an endomorphism of $V = V_0 \oplus V_1$ by setting $\theta(V_0) = 0$.
By definition $G$ acts on $\{1,2,\ldots, n\}$. Let $H = \mathrm{Stab}_{G}(1)$. Suppose that
$$\theta(e_1) = a e_1 + \sum_{i=2}^n b_i e_i$$
where $a \in \mathbb{C}$ and $b_i \in \mathbb{C}$ for each $i \in \{2,\ldots, n\}$.
Since $h \cdot 1 = 1$ for each $h \in H$, and $\theta$ is a $\mathbb{C}G$-homomorphism, the vector
$$\sum_{i=2}^n b_i e_i \in V_1 $$
is $H$-invariant. But $H$ has a single orbit on $\{2,\ldots,n\}$ because $G$ is $2$-transitive. Therefore $b_i$ is constant for $i \in \{2,\ldots,n\}$. Let $b$ be the common value.
To complete the proof, pick $g_1, \ldots, g_n \in G$ such that $g_i \cdot 1 = i$ for each $i \in \{1,2,\ldots, n\}$. We have
$$ 0 = \theta\bigl( \sum_{i=1}^n e_i \bigr) = \theta\bigl( \sum_{i=1}^n g_i \cdot e_1 \bigr)
= \sum_{i=1}^n g_i \cdot \theta(e_1). $$
Comparing the coefficient of $e_1$ on both sides gives $a+(n-1)b = 0$. So $\theta$ is determined by $a$ and, as claimed, $\mathrm{dim}\ \mathrm{End}_{\mathbb{C}G} (V_1) = 1$.
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 $\pi$.
Best Answer
There are $|H|$ elements with $|G/H|$ fixed points, and thus trace $|G/H|$, and all other elements have no fixed points and are thus traceless. Thus, by the orthogonality relations, the multiplicity of the trivial representation $1$ in $\rho$ is
$$ \frac1{|G|}\sum_g\chi_1(g)\chi_\rho(g)=\frac1{|G|}\left(|H|\frac{|G|}{|H|}\right)=1. $$