This is from Tu's book in smooth manifold. Let $V$ be a vector space, and let us define $S_n$ acting on $\Lambda^n(V^*)$ as follows:
For $\sigma \in S_n$ and $f \in \Lambda^n(V^*)$, $\sigma f(v_1, \cdots, v_n) = f(v_{\sigma(1)}, \cdots, v_{\sigma(n)})$.
Then, there is a lemma:
Lemma: For $\sigma, \tau \in S_n$, $\tau(\sigma f) = (\tau\sigma)f$.
Proof: For $v_1, \cdots, v_n \in V$, \begin{align*} \tau(\sigma f) (v_1, \cdots, v_k) &= (\sigma f)(v_{\tau(1)}, \cdots, v_{\tau(k)}) \\ &= (\sigma f)(w_1, \cdots, w_k) \quad \quad \text{by letting $w_i = v_{\tau(i)})$} \\ &= f(w_{\sigma(1)}, \cdots, w_{\sigma(k)}) \\ &= f(v_{\tau(\sigma(1))}, \cdots, v_{\tau(\sigma(n))}) = f(v_{\tau\sigma(1)}, \cdots, v_{\tau\sigma(n)}) \\ &= (\tau\sigma) f(v_1, \cdots, v_k). \end{align*}
The proof is perfectly understandable. However, when I tried to apply the theorem holds for specific case, I am not sure what I am doing wrong. Say $n = 4$ and $\tau = (12)(34)$ and $\sigma = (23)$. Then, $\tau[1,2,3,4] = [2,1,4,3]$ and $\sigma[1,2,3,4] = [1,3,2,4]$. Also, $\tau\sigma[1,2,3,4] = [3,1,4,2]$. Hence, $(\tau\sigma)f(v_1, \cdots, v_4) = f(v_3, v_1, v_4, v_2)$.
If I define $(w_1, w_2, w_3, w_4) = (v_{\tau(1)}, \cdots, v_{\tau(4)})$, then $(w_1, w_2, w_3, w_4) = (v_2, v_1, v_4, v_3)$. Hence,
\begin{align*} \tau(\sigma f)(v_1, \cdots, v_4) &= (\sigma f)(v_{\tau(1)}, \cdots, v_{\tau(4)}) \\ &= (\sigma f)(w_1, \cdots, w_4) \\ &= f (w_{\sigma(1)}, \cdots, w_{\sigma(4)}) \\ &= f (w_1, w_3, w_2, w_4) \\ &= f (v_2, v_4, v_1, v_3) \end{align*}
This is not the same result as $f(v_3, v_1, v_4, v_2)$. What am I doing wrong here?
Your mistake is the claim that $\tau\sigma[1,2,3,4]=[3,1,4,2]$. That composition works like this: \begin{align*} \tau\sigma(1) &= \tau(\sigma(1)) = \tau(1)=2;\\ \tau\sigma(2) &= \tau(\sigma(2)) = \tau(3)=4;\\ \tau\sigma(3) &= \tau(\sigma(3)) = \tau(2)=1;\\ \tau\sigma(4) &= \tau(\sigma(4)) = \tau(4)=3. \end{align*} Therefore, $\tau\sigma[1,2,3,4] = [2,4,1,3]$, and there's no contradiction.
Here's a systematic way to think about these operations. Let $V$ be a vector space. An element of $V^k$ is an ordered $k$-tuple, which is actually a function from $\{1,\dots,k\}$ to $V$. On the other hand, a $k$-form $f$ is a function from $V^k$ to $\mathbb R$.
If $v = (v_1,\dots,v_k) \in V^k$ and $\sigma$ is a permutation of $\{1,\dots,k\}$, then the $k$-tuple $(v_{\sigma(1)},\dots,v_{\sigma(k)})$ is actually the composition $v \circ σ: \{1,\dots,k\} \to V$. Therefore, by definition, $σf$ is the map from $V^k$ to $\mathbb R$ given by $$ \sigma f(v) = f (v \circ \sigma). $$ It follows that $$ \sigma\tau f (v) = f (v \circ \sigma\circ\tau) = \tau f(v \circ \sigma) = \sigma(\tau f)(v). $$ Does that make sense?