If $P_j$ is a projector of the form $P_j=I_X\otimes \left|j\right>\left<j\right|$ then given $\left|x\right>\in X$ and operator $A_i: X\rightarrow X$ is it true that, for if $\{\left|i\right>\}$ forming an orthonormal basis
$$\sum_{j}P_{j}\sum_{i}A_i\left|x\right>\left|i\right>=\sum_{i}P_iA_i\left|x\right>\left|i\right>$$
I've convinced myself of this with
$$\begin{aligned} \sum_{j}P_{j}\sum_{i}A_i\left|x\right>\left|i\right>&=\sum_{j,i}P_{j}A_i\left|x\right>\left|i\right>\\ &=\sum_{j,i}\left(I_X\otimes \left|j\right>\left<j\right|\right)A_i\left|x\right>\left|i\right>\\ &=\sum_{j,i}\left(I_XA_i\left|x\right>\right)\otimes \left|j\right>\left<j\right|\left|i\right>\\ &=\sum_{j=i}A_i\left|x\right>\otimes \left<j\vert i\right>\left|j\right>+\sum_{j\neq i}A_i\left|x\right>\otimes \left<j\vert i\right>\left|j\right>\\ &=\sum_{j=i}A_i\left|x\right>\otimes \left<j\vert i\right>\left|j\right>\\ &=\sum_{i}A_i\left|x\right>\otimes \left<i\vert i\right>\left|i\right>\\ &=\sum_{i}\left(I_XA_i\left|x\right>\right)\otimes \left|i\right>\left<i\right|\left|i\right>\\ &=\sum_{i}\left(I_X\otimes \left|i\right>\left<i\right|\right)A_i\left|x\right>\left|i\right>\\ &=\sum_{i}P_{i}A_i\left|x\right>\left|i\right>\,\square\\ \end{aligned}$$
but I'm not that proficient at matrix operations or summation tricks and fear I may have done violence ot both.
Yes, if $x\in X$ and if $|x\rangle\,|i\rangle$ means $|x\rangle \otimes |i\rangle$, then your computation is correct.