Suppose $X$ is an $n$-dimensional manifold, and consider the Bockstein homomorphism $\beta \colon H_k(X; \mathbb{Z}_2) \to H_{k-1}(X; \mathbb{Z})$ induced by the sequence $$0 \to \mathbb{Z} \xrightarrow{\cdot 2} \mathbb{Z} \xrightarrow{\mathrm{mod \, 2}} \mathbb{Z}_2 \to 0.$$
It seems well-known that $(\mathrm{mod\,2}) \circ \beta :H_n(X;\mathbb{Z}_2) \to H_{n-1}(X;\mathbb{Z}_2)$ sends the mod 2 fundamental class $[X]_2$ to an $(n-1)$-cycle whose mod 2 Poincare dual is $w_1(X)$. Frankly, I don't see it.
If $X$ is orientable, we can use the fact that Poincare dual and (the dual) Bockstein commute (the cap product is a chain map up to sign if composed with a cycle), that is
$$\require{AMScd} \begin{CD} H_n(X;\mathbb{Z}_2) @>{\beta}>> H_{n-1}(X;\mathbb{Z}) @>\mathrm{mod \, 2}>> H_{n-1}(X;\mathbb{Z}_2) \\ @V{\approx}V{\mathrm{PD}_2}V @V{\approx}V{\mathrm{PD}}V @V{\approx}V{\mathrm{PD}_2}V \\ H^0(X;\mathbb{Z}_2) @>{\beta}>> H^1(X;\mathbb{Z}) @>\mathrm{mod \, 2}>> H^1(X;\mathbb{Z}_2) \end{CD}$$
and then we apply the fact that $(\mathrm{mod\,2}) \circ \beta = \mathrm{sq}^1$ which is zero if we apply it to the Poincare dual of $[X]_2$. But in that case $w_1(X)$ vanishes anyways. How do I show this in the non-orientable case? It feels like I oversee something very obvious here...