Given four matrices $A, \widetilde{A}, B, \widetilde{B} \in \mathbb{R}^{n\times d}$, if $A^{\top} A \approx_{\epsilon} \widetilde{A}^{\top} \widetilde{A}$, $B^{\top} B \approx_{\epsilon} \widetilde{B}^{\top} \widetilde{B}$, do we have \begin{align*} (A+B)^{\top} \cdot (A+B) \approx_{10\epsilon} (\widetilde{A}+\widetilde{B})^{\top} \cdot (\widetilde{A}+\widetilde{B}) ? \end{align*}
For two square matrices $C, \widetilde{C}$, we say $C \approx_{\epsilon} \widetilde{C}$ if \begin{align*} (1-\epsilon) \widetilde{C} \preceq C \preceq (1+\epsilon) \widetilde{C} \end{align*}
No. The problem is that if $X^\top X=P$, there is great freedom to choose $X$. In fact, if $X^\top X$ is equal to $P$, so is $Y^\top Y$ whenever $Y=QX$ for some real orthogonal matrix $Q$. In particular, it can happen that $(A+A)^\top(A+A)$ is not close to $(A+QA)^\top(A+QA)$. E.g. when $A=I_2$ and $Q=\pmatrix{0&1\\ 1&0}$, $$ (A+A)^\top(A+A)-(A+QA)^\top(A+QA) =\pmatrix{2&-2\\ -2&2} $$ is indefinite. Therefore, when $\epsilon$ is small, neither $(A+A)^\top(A+A)\approx_{10\epsilon}(A+QA)^\top(A+QA)$ nor $(A+QA)^\top(A+QA)\approx_{10\epsilon}(A+A)^\top(A+A)$ is true.