An application of the Pressing Down Lemma (Part 1)

For each $\sigma$ and $\alpha$ let $P = W_\sigma - \alpha$ which is still stationary. Now for each $f^* \in P$, let $g_\sigma(f^*)$ be $f(n)$ where $n$ is least such that $f(n) \ge \alpha$. Now using pressing down lemma and the pigeonhole principle, you can find some fixed $n$ and some fixed $\gamma \ge \alpha$ such that $\{ f^* \in P: f(n) = \gamma\}$ is stationary. Now if $n\le |\sigma|$, you are done. Else to fill the gap between $|\sigma|$ and $n$ you can use repeated applications of the pressing down lemma, to get the desired $\theta$.

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EDIT: [This edit will try to complete the below answer you provided.][Disclaimer: I am using your notation.]

As in your answer, let $S = \{f^* \in P: f(m) = \gamma\}$ and suppose $m \gt |\sigma|$. By the fact that $P \subseteq W_\sigma$, we have $f||\sigma| = \sigma$, for any $f^* \in S$. First we inductively choose a finite sequence of stationary sets $\langle S_0, \dots, S_{m-|\sigma|-1}\rangle$ and a finite sequence of ordinals $\langle \beta_0, \dots, \beta_{m-|\sigma|-1}\rangle$, such that $S_0 \subseteq S$, $S_{i+1} \subseteq S_i$, for $i \lt m-|\sigma|-1$. Also we make sure that for each $f^* \in S_i$, $f(i+|\sigma|) = \beta_i$.

This can be done easily, using the pressing down lemma. For the base case $i = 0$, consider $g(f^*) = f(|\sigma|)$ and by the pressing down lemma you have some stationary $S_0 \subseteq S$ and some ordinal $\beta_0$ such that $g"S_0 = \{\beta_0\}$. At the $i$th step just look at $g(f^*) = f(i+|\sigma|)$, and construct $S_i$ and $\beta_i$ as above.

So we wish to build a $\theta \in \chi^{m+1}$ that satisfies the conditions in the question. First let $\theta||\sigma| = \sigma$ and $\theta(m) = \gamma$. Now for $|\sigma| \le i \lt m$, let $\theta(i) = \beta_{i-|\sigma|}$. Now you can see that $W_\theta$ is stationary as it contains $S_{m-|\sigma|-1}$. And also because of $\gamma$ you have $\theta \not \in \cup_{n\in\omega} \alpha^n$.