Let $C_iU_j$ denote the controlled-$U$ gate with qubit $i$ as the control and qubit $j$ as the target. The key observation is that the CZ gate, whose action on the computational basis is $$ \begin{align} |00\rangle&\mapsto|00\rangle\\ |01\rangle&\mapsto|01\rangle\\ |10\rangle&\mapsto|10\rangle\\ |11\rangle&\mapsto-|11\rangle\\ \end{align}\tag1 $$ is symmetric under the exchange of qubits. In other words, $$ C_1Z_2=C_2Z_1.\tag2 $$ Therefore, $$ \begin{align} (H\otimes H)\circ C_1X_2\circ(H\otimes H)&=(H\otimes I)\circ C_1Z_2\circ(H\otimes I)\tag3\\ &=(H\otimes I)\circ C_2Z_1\circ(H\otimes I)\tag4\\ &=C_2X_1\tag5 \end{align} $$ where we first used $HXH=Z$, then $(2)$ and then $HZH=X$ again. $\square$

The key observation is that the CZ gate, whose action on the computational basis is $$ \begin{align} |00\rangle&\mapsto|00\rangle\\ |01\rangle&\mapsto|01\rangle\\ |10\rangle&\mapsto|10\rangle\\ |11\rangle&\mapsto-|11\rangle\\ \end{align}\tag1 $$ is symmetric under the exchange of qubits. In other words, $$ C_1Z_2=C_2Z_1\tag2 $$ where $C_iU_j$ denotes the controlled-$U$ gate with qubit $i$ as the control and qubit $j$ as the target. Therefore, $$ \begin{align} (H\otimes H)\circ C_1X_2\circ(H\otimes H)&=(H\otimes I)\circ C_1Z_2\circ(H\otimes I)\tag3\\ &=(H\otimes I)\circ C_2Z_1\circ(H\otimes I)\tag4\\ &=C_2X_1\tag5 \end{align} $$ where we first used $HXH=Z$, then $(2)$ and then $HZH=X$ again. $\square$