How to decompose linear transformation into "lerp-able" products of rotation and scaling transforms? Given two linear transforms in 3D as a matrices $A_0,A_1 \in \mathbb{R}^{3 \times 3}$. I can lerp between them for any "time", $t \in [0,1]$ using an expression such as$^1$
$$
A_t = \exp\left( (1-t) \log(A_0) + t \log(A_1)\right).
$$
This gives pleasing interpolations, such as in this example where $A_0 = I$ and $A_1 = \begin{bmatrix}1&-1&0\\0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$:

Now, consider that I'm not allowed to store or interpolate linear transformations directly. I must decompose each transform into a product of rotations and diagonal matrices (a.k.a., non-uniform scaling transforms). I.e., $A_0 = R_0^1 S_0^1 \dots R_0^n S_0^n$ and $A_1 = R_1^1 S_1^1 \dots R_1^n S_1^n$.
Interpolations will be conducted by slerping rotations and linearly interpolating diagonal matrix entries.
How can I emulate the linear interpolation above under these conditions?
As an example of a poor solution, consider decomposing the input transformations using singular value decomposition:
$$
A = U S V^\top \\
A = \underbrace{\det{(U)}\ U}_{R^1} \  \underbrace{\det{(U)} \ S \ \det{(V)}}_{S^1} \ \underbrace{\det{(V)}\  V^\top}_{R^2} \\
A = R^1 S^1 R^2
$$
By construction, $R^1,R^2 \in SO(3)$ and $S^1$ is diagonal.
Slerping and lerping these transforms for the problem above via:
$$
A_t = \text{slerp}\left(R^1_0,R^1_1,t\right)\ \left((1-t) S^1_0 + t S^1_1\right)\ \text{slerp}\left(R^2_0,R^2_1,t\right) 
$$
produces an interpolation with the correct end-point values but wild behavior in between:

Is it fundamentally impossible to get a better interpolation? If not, would using more than two auxiliary rotations per transform help tame this interpolation?
Typically, I have a sequence of transforms $A_0,A_1,A_2, \dots$. So, another phrasing would be, given a decomposition for $A_0$ (i.e., all $R^i_0$ and $S^i_0$ are given) what values for the decomposition of $A_1$ would create the least-motion/least-work interpolation?
 A: Another method to interpolate from $I$ to a matrix $A$ with $\det(A) > 0$: let $A = USV^T$ be an SVD. We can write
$$
A = \underbrace{UV^T}_{R_1}\ \  \underbrace{VSV^T}_B,
$$
noting that $R_1$ is guaranteed to be a rotation when $\det(A) > 0$.  Notably, $B = VSV^T$ is the spectral decomposition of a positive definite matrix. If $V$ is a rotation, then we're done. Otherwise, as below, we can take
$$
B = \underbrace{VD_\epsilon}_{R_2} \ S\ \underbrace{D_{\epsilon} V^T}_{R_3}.
$$
I believe that "slerping" preserves the property that $R_2$ and $R_3$ are inverses, which I suspect means that the choice of $D_\epsilon$ used here has no effect on the overall transformation.

An improvement to the approach below: denote
$$
D_{\epsilon}^x = \pmatrix{\epsilon &0&0\\0&1&0\\0&0&1},\quad
D_{\epsilon}^{y} = \pmatrix{1&0&0\\0&\epsilon&0\\0&0&1}, \quad 
D_{\epsilon}^{z} = \pmatrix{1&0&0\\0&1&0\\0&0&\epsilon}.
$$
For each choice of $D_{\epsilon} \in \{D_{\epsilon}^x,D_\epsilon^y,D_\epsilon^z,\epsilon I\}$, the product
$$
A = U S V^T = \underbrace{U D_{|U|}}_{R^1}
\underbrace{D_{|U|}SD_{|V|}}_{S^1}
\underbrace{D_{|V|}V^T}_{R^2}
$$
is a suitable decomposition. I suspect that the "wildness" of the rotation is determined by the angle by which $R^1$ and $R^2$ rotate; a lower angle means a more parsimonious interpolation.
With that in mind, I suspect that a good heuristic for the best candidate $D_{\epsilon} \in \{D_{\epsilon}^x,D_\epsilon^y,D_\epsilon^z,\epsilon I\}$ is to choose the candidate for which the sum of the angle-cosines is maximized, i.e. the candidate for which
$$
\operatorname{tr}(UD_{|U|}) + \operatorname{tr}(D_{|V|}V^T)
$$
is maximized.

I suspect that using the SVD is a reasonable approach here, but that the particular way that you have chosen to make $U$ and $V$ into rotation matrices (i.e. by multiplying by their determinants) has led to this issue. I suggest the following alternative.
Define
$$
D_{\epsilon} = \pmatrix{1&0&0\\0&1&0\\0&0&\epsilon}.
$$
Decompose $A$ as the product
$$
A = U S V^T = \underbrace{U D_{|U|}}_{R^1}
\underbrace{D_{|U|}SD_{|V|}}_{S^1}
\underbrace{D_{|V|}V^T}_{R^2}.
$$
