Singular value decomposition
Start with a matrix with $m$ rows, $n$ columns, and rank $\rho$,
$$
\mathbf{A}\in\mathbb{C}^{m\times n}_{\rho}
$$
which has the singular value decomposition
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]^{*}
%
$$
where the color denotes $\color{blue}{range}$ spaces and $\color{red}{null}$ spaces. The dimensions of the domain matrices are
$$
%
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}}
\in \mathbb{C}^{m\times \rho}, \quad
%
\color{red}{\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\in \mathbb{C}^{m \times m - \rho}, \quad
%
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\in \mathbb{C}^{n\times \rho}, \quad
%
\color{red}{\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\in \mathbb{C}^{n\times n - \rho}.
$$
The domain matrices are unitary:
$$
\begin{align}
\mathbf{U}\mathbf{U}^{*} &= \mathbf{U}^{*}\mathbf{U} = \mathbf{I}_{m} \\
\mathbf{V}\mathbf{V}^{*} &= \mathbf{V}^{*}\mathbf{V} = \mathbf{I}_{n}
\end{align}
$$
The dimensions of the singular value matrices are
$$
%
\Sigma \in \mathbb{R}^{m\times n}, \quad
%
\mathbf{S}
\in \mathbb{R}^{\rho\times \rho}.
$$
The hermitian conjugate is constructed according to
$$
\mathbf{A}^{*} =
\mathbf{V} \, \Sigma^{\mathrm{T}} \, \mathbf{U}^{*} =
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]^{*}
%
$$
where $\Sigma^{\mathrm{T}}\in \mathbb{R}^{n\times m}$.
The Moore-Penrose pseudoinverse is constructed according to
$$
\mathbf{A}^{\dagger} =
\mathbf{V} \, \Sigma^{\dagger} \, \mathbf{U}^{*} =
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]^{*}
%
$$
where $\Sigma^{\dagger}\in \mathbb{R}^{n\times m}$.
The product matrix rules you stated always hold:
$$
\begin{align}
%
\mathbf{A} \mathbf{A}^{*} &=
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)^{*}
%
=
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)
%
\left(
\mathbf{V} \, \mathbf{\Sigma}^{\mathrm{T}} \, \mathbf{V}^{*}
\right) \\
%
\mathbf{A}^{*} \mathbf{A} &=
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)^{*}
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right)
%
=
%
\left(
\mathbf{V} \, \mathbf{\Sigma}^{\mathrm{T}} \, \mathbf{V}^{*}
\right)
%
\left(
\mathbf{U} \, \mathbf{\Sigma} \, \mathbf{V}^{*}
\right) \\
%
\end{align}
$$
Examples follow.
Square, full rank $m = n = \rho$
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \mathbf{S} \right]
%
\left[ \begin{array}{c}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]^{*}
%
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{*}\mathbf{A} &= \color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}^{*} \\
%
\mathbf{A}\mathbf{A}^{*} &= \color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}^{*}
%
\end{align}
$$
Tall, full column rank $n = \rho$, $m \ge n$
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]
%
\left[ \begin{array}{c}
\mathbf{S} \\ \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]^{*}
%
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{*}\mathbf{A} &= \color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}}^{*} \\
%
\mathbf{A}\mathbf{A}^{*} &=
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} &
\color{red} {\mathbf{U}_{\mathcal{N}\left(\mathbf{A}^{*}\right)}}
\end{array} \right]^{*}
%
\end{align}
$$
Wide, full row rank $m = \rho$, $n \ge m$
$$
\mathbf{A} =
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} =
%
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]^{*}
%
$$
The product matrices are
$$
\begin{align}
%
\mathbf{A}^{*}\mathbf{A} &=
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]
%
\left[ \begin{array}{cc}
\mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{cc}
\color{blue}{\mathbf{V}_{\mathcal{R}\left(\mathbf{A}^{*}\right)}} &
\color{red} {\mathbf{V}_{\mathcal{N}\left(\mathbf{A}\right)}}
\end{array} \right]^{*} \\
%
\mathbf{A}\mathbf{A}^{*} &=
%
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} \,
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{U}_{\mathcal{R}\left(\mathbf{A}\right)}} \\
%
\end{align}
$$
For the hermitian matrix,
$$
\begin{align}
\mathbf{A} &= \mathbf{A}^{*} \\
\mathbf{U} \, \Sigma \, \mathbf{V}^{*} &=
\mathbf{V} \, \Sigma \, \mathbf{U}^{*}
\end{align}
$$
because in this case $\Sigma = \Sigma^{\mathrm{T}}$.
Best Answer
Covariance matrices are positive-semidefinite, and PSD matrices have unique PSD square roots (given by taking the unique nonnegative square root of each eigenvalue). This means that $V^T A V$ is the unique PSD square root of $V^T A^2 V$. We have $V^T A^2 V = \Sigma^2$ and the unique PSD square root of $\Sigma^2$ is $\Sigma$, so $V^T A V = \Sigma$.
We can give an alternative analysis in terms of eigenspaces as follows. Consider some eigenspace $E_{\lambda}$ of $A^2$. By definition we have $A^2 v = \lambda v$ for all $v \in E_{\lambda}$. Since $A$ commutes with $A^2$, it restricts to a map $A : E_{\lambda} \to E_{\lambda}$ which squares to $\lambda$. You are correct that in general it does not follow that $A$ acts by a scalar (and good job spotting this possibility!), but if $\lambda \ge 0$ and $A$ is PSD then $A$ must act by $\sqrt{\lambda}$. This is because $\sqrt{\lambda}$ is the only possible eigenvalue of $A$ here (and $A$ is diagonalizable by the spectral theorem).