Consider the following:
$$ \mathbf{A}^\text{T}\mathbf{A}+\mathbf{B}^\text{T}\mathbf{B} \tag{1} $$ where $\mathbf{A}$ and $\mathbf{B}$ are $n\times m$ matrices. Is it possible to write this in the form
$$ \left(\mathbf{X}+\mathbf{Y}\right)^\text{T}\left(\mathbf{X}+\mathbf{Y}\right) \tag{2} $$ via some transformation? I.e
$$ \mathbf{X}^\text{T}\mathbf{X}+\mathbf{Y}^\text{T}\mathbf{Y}+\mathbf{X}^\text{T}\mathbf{Y}+\mathbf{Y}^\text{T}\mathbf{X}=\mathbf{A}^\text{T}\mathbf{A}+\mathbf{B}^\text{T}\mathbf{B} \tag{3} $$
What I'm hoping to achieve is to preserve $\left(\mathbf{X}+\mathbf{Y}\right)$ as an $n\times m$ matrix. Rather than Cholesky decompose $\mathbf{A}^\text{T}\mathbf{A}+\mathbf{B}^\text{T}\mathbf{B} $ directly and end up with an $m\times m$ matrix.
UPDATE: An attempt. Let
$$ \mathbf{A} = \pmb{\alpha}\mathbf{Y}+\pmb{\beta}\mathbf{X} \tag{4} $$ And $$ \mathbf{B} = \pmb{\gamma}\mathbf{Y}+\pmb{\delta}\mathbf{X} \tag{5} $$ Then $$ \begin{align} \mathbf{A}^\text{T}\mathbf{A} & = \left( \pmb{\alpha}\mathbf{Y}+\pmb{\beta}\mathbf{X}\right)^\text{T} \left( \pmb{\alpha}\mathbf{Y}+\pmb{\beta}\mathbf{X}\right) \\ & = \mathbf{Y}^\text{T}\pmb{\alpha}^\text{T}\pmb{\alpha}\mathbf{Y} + \mathbf{X}^\text{T}\pmb{\beta}^\text{T}\pmb{\beta}\mathbf{X} + \mathbf{Y}^\text{T}\pmb{\alpha}^\text{T}\pmb{\beta}\mathbf{X}+ \mathbf{X}^\text{T}\pmb{\beta}^\text{T}\pmb{\alpha}\mathbf{Y} \end{align} \tag{6} $$ Similarly:
$$ \begin{align} \mathbf{B}^\text{T}\mathbf{B} & = \left( \pmb{\gamma}\mathbf{Y}+\pmb{\delta}\mathbf{X}\right)^\text{T} \left( \pmb{\gamma}\mathbf{Y}+\pmb{\delta}\mathbf{X}\right) \\ & = \mathbf{Y}^\text{T}\pmb{\gamma}^\text{T}\pmb{\gamma}\mathbf{Y} + \mathbf{X}^\text{T}\pmb{\delta}^\text{T}\pmb{\delta}\mathbf{X} + \mathbf{Y}^\text{T}\pmb{\gamma}^\text{T}\pmb{\delta}\mathbf{X}+ \mathbf{X}^\text{T}\pmb{\delta}^\text{T}\pmb{\gamma}\mathbf{Y} \end{align} \tag{7} $$ Therefore
\begin{align} \mathbf{A}^\text{T}\mathbf{A} + \mathbf{B}^\text{T}\mathbf{B} & = \left( \pmb{\alpha}\mathbf{Y}+\pmb{\beta}\mathbf{X}\right)^\text{T} \left( \pmb{\alpha}\mathbf{Y}+\pmb{\beta}\mathbf{X}\right) \\ & = \mathbf{Y}^\text{T}\left(\pmb{\alpha}^\text{T}\pmb{\alpha}+\pmb{\gamma}^\text{T}\pmb{\gamma}\right)\mathbf{Y} + \mathbf{X}^\text{T}\left(\pmb{\beta}^\text{T}\pmb{\beta} + \pmb{\delta}^\text{T}\pmb{\delta}\right)\mathbf{X} + \mathbf{Y}^\text{T}\left(\pmb{\alpha}^\text{T}\pmb{\beta}+\pmb{\gamma}^\text{T}\pmb{\delta}\right)\mathbf{X}+ \mathbf{X}^\text{T}\left(\pmb{\beta}^\text{T}\pmb{\alpha} + \pmb{\delta}^\text{T}\pmb{\gamma}\right)\mathbf{Y} \tag{8} \end{align}
Comparing this with eqn 2, then we have the following conditions:
\begin{align} & \pmb{\alpha}^\text{T}\pmb{\alpha}+\pmb{\gamma}^\text{T}\pmb{\gamma} = \pmb{1}\\ & \pmb{\beta}^\text{T}\pmb{\beta} + \pmb{\delta}^\text{T}\pmb{\delta} = \pmb{1}\\ & \pmb{\alpha}^\text{T}\pmb{\beta}+\pmb{\gamma}^\text{T}\pmb{\delta} = \pmb{1}\\ & \pmb{\beta}^\text{T}\pmb{\alpha} + \pmb{\delta}^\text{T}\pmb{\gamma}= \pmb{1}\\ \tag{9} \end{align}
From equation 4 we have that: \begin{align} \mathbf{X} = \pmb{\beta}^{-1}\left(\mathbf{A} - \pmb{\alpha}\mathbf{Y}\right) \tag{10} \end{align} Substituting into equation 5:
$$ \mathbf{B} = \pmb{\gamma}\mathbf{Y}+\pmb{\delta}\pmb{\beta}^{-1}\left(\mathbf{A} - \pmb{\alpha}\mathbf{Y}\right) \tag{11} $$ Expanding:
$$ \mathbf{B} = \pmb{\gamma}\mathbf{Y}+\pmb{\delta}\pmb{\beta}^{-1}\mathbf{A} - \pmb{\delta}\pmb{\beta}^{-1}\pmb{\alpha}\mathbf{Y} \tag{12} $$ Writing in terms of $\mathbf{Y}$ we find that
$$ \mathbf{Y} = \left(\pmb{\gamma}- \pmb{\delta}\pmb{\beta}^{-1}\pmb{\alpha}\right)^{-1}\left(\mathbf{B}-\pmb{\delta}\pmb{\beta}^{-1}\mathbf{A}\right) \tag{13} $$ substituting into equation 10:
$$ \mathbf{X} = \pmb{\beta}^{-1}\left(\mathbf{A} - \pmb{\alpha}\left(\pmb{\gamma}- \pmb{\delta}\pmb{\beta}^{-1}\pmb{\alpha}\right)^{-1}\left(\mathbf{B}-\pmb{\delta}\pmb{\beta}^{-1}\mathbf{A}\right)\right) \tag{14} $$
And hence
$$ \mathbf{X} + \mathbf{Y} = \left(\pmb{\gamma}- \pmb{\delta}\pmb{\beta}^{-1}\pmb{\alpha}\right)^{-1}\left(\mathbf{B}-\pmb{\delta}\pmb{\beta}^{-1}\mathbf{A}\right) + \pmb{\beta}^{-1}\left(\mathbf{A} - \pmb{\alpha}\left(\pmb{\gamma}- \pmb{\delta}\pmb{\beta}^{-1}\pmb{\alpha}\right)^{-1}\left(\mathbf{B}-\pmb{\delta}\pmb{\beta}^{-1}\mathbf{A}\right)\right) \tag{15} $$ subject to \begin{align} & \pmb{\alpha}^\text{T}\pmb{\alpha}+\pmb{\gamma}^\text{T}\pmb{\gamma} = \pmb{1}\\ & \pmb{\beta}^\text{T}\pmb{\beta} + \pmb{\delta}^\text{T}\pmb{\delta} = \pmb{1}\\ & \pmb{\alpha}^\text{T}\pmb{\beta}+\pmb{\gamma}^\text{T}\pmb{\delta} = \pmb{1}\\ & \pmb{\beta}^\text{T}\pmb{\alpha} + \pmb{\delta}^\text{T}\pmb{\gamma}= \pmb{1}\\ \tag{16} \end{align} Is there a way to find these four matrices?
Note that your matrix $C=A^TA+B^TB$ is guaranteed to be a symmetric matrix (in fact positive definite as well). Note that the rank $r$ of your matrix obeys $r\leq \min(n,m)$. W.l.o.g assume $r=n$. You can write the eigen-decomposition as $$C=\sum_{i=1}^{n}\lambda_i u_i u_i^T$$ where $u_i$ are the $m\times 1$ eigenvectors. Convince yourself that you can write this as $$C=\hat{U}\hat{\Sigma}\hat{U}^T$$ where $\hat{U}$ is the $m\times n$ matrix whose columns are the eigen-vectors and $\hat{\Sigma}$ is the $n \times n$ diagonal with the eigen values. Now, your requirement becomes straightforward. Define $R=\hat{U}\hat{\Sigma}^{1/2}$ so that $C=R^TR$ and $R$ is $n \times m$ matrix.