You seem to be using $\partial$ to mean both the connecting homomorphism and the boundary map of chain complexes. While related, they are not exactly the same. There are also several connecting homomorphisms in play here associated to the short exact sequences of complexes
\begin{equation}
C_\bullet(A; ?) \to C_\bullet(X; ?) \to C_\bullet(X,A; ?)
\end{equation}
\begin{equation}
C_\bullet(??; G') \to C_\bullet(??; G) \to C_\bullet(??; G''),
\end{equation}
where $? \in \{G', G, G''\}$ in the first line and $?? \in \{A, X, (X,A)\}$ in the second, and you should take care not to confuse them.

Anyway, here is a hands-on way of showing that the Bockstein commutes with the connecting homomorphism of pairs of spaces. Assuming you've already shown that the chain level definitions are well-defined, you want to show that the following diagram "commutes":
\begin{equation}
\begin{array}{ccccccc}
C_n(X,A;G'') & \leftarrow & C_n(X,A;G) & \rightarrow & C_{n-1}(X,A;G) & \leftarrow & C_{n-1}(X,A;G') \\
\uparrow & & & & & & \uparrow \\
C_n(X;G'') & & & & & & C_{n-1}(X;G') \\
\downarrow & & & & & & \downarrow \\
C_{n-1}(X;G'') & & & & & & C_{n-2}(X;G') \\
\uparrow & & & & & & \uparrow \\
C_{n-1}(A;G'') & \leftarrow & C_{n-1}(A;G) & \rightarrow & C_{n-2}(A;G) & \leftarrow & C_{n-2}(A;G')
\end{array}
\end{equation}

For example, the top row should be read as "take a cycle in $C_n(X,A;G'')$, lift it to a chain in $C_n(X,A;G)$ using the short exact sequence of groups, apply the boundary map to get an element in $C_{n-1}(X,A;G)$, then use the short exact sequence of groups again to get an cycle in $C_{n-1}(X,A;G')$". Upon passing to homology, this is the definition of the Bockstein homomorphism. The other sides of the diagram should be interpreted similarly.

To show that the diagram "commutes", simply note that it is built out of smaller squares:
\begin{equation}
\begin{array}{ccccccc}
C_n(X,A;G'') & \leftarrow & C_n(X,A;G) & \rightarrow & C_{n-1}(X,A;G) & \leftarrow & C_{n-1}(X,A;G') \\
\uparrow & & \uparrow & & \uparrow & & \uparrow \\
C_n(X;G'') & \leftarrow & C_n(X;G) & \rightarrow & C_{n-1}(X;G) & \leftarrow & C_{n-1}(X;G') \\
\downarrow & & \downarrow & & \downarrow & & \downarrow \\
C_{n-1}(X;G'') & \leftarrow & C_{n-1}(X;G) & \rightarrow & C_{n-2}(X;G) & \leftarrow & C_{n-2}(X;G') \\
\uparrow & & \uparrow & & \uparrow & & \uparrow \\
C_{n-1}(A;G'') & \leftarrow & C_{n-1}(A;G) & \rightarrow & C_{n-2}(A;G) & \leftarrow & C_{n-2}(A;G')
\end{array}
\end{equation}

You should check that each of the smaller squares commute. Therefore, the large outer square "commutes" too, and this shows that the Bockstein and the connecting homomorphism commutes.

**EDIT:** Following the discussion in the comments, here is a simplified toy example that might make things clearer. Consider the map of chain complexes:
\begin{equation}
\begin{array}{ccc}
0 & \rightarrow & \mathbb{Z} \\
\downarrow & & \downarrow \\
\mathbb{Z} & \xrightarrow{f} & \mathbb{Z} \\
\downarrow & & \downarrow \\
\mathbb{Z} & \rightarrow & 0
\end{array}
\end{equation}
where the maps $\mathbb{Z} \to \mathbb{Z}$ are all identities and everything else is $0$. It's obvious that $\partial^2 = 0$ for both the chain complexes on the left and right, but if you start at the upper right corner, apply $\partial$, then $f^{-1}$, and then $\partial$, you would get the identity map, not zero.

## Best Answer

This is an instance of the general fact that the Steenrod operation $\operatorname{Sq}^i$ squares elements of degree $i$: $\operatorname{Sq}^i(a) = a \cup a$ for $|a|=i.$ It is known that $\operatorname{Sq}^1 = \beta$.

Now, if you want to see it directly, then yes: first of all, you note that $\beta$ maps the generator to the generator: this is seen directly from the definition of $\beta$ by writing chain complexes and tracing the image of the generator $a$ of $H^1.$ Then knowing that the cup product $a \cup a$ if nontrivial will mean that it is equal to $\beta(a).$ Nontrivialoty of the cup product on the projective space is a classical (and pivotal) result, and there are many ways to prove it. The one you suggest is one of the least painful. You note that cup product corresponds via Poicare duality to intersecting things and use the geometry of $\mathbb{R}P^2$ to deduce the result.