\newcommand{\ann}{\operatorname{ann}}
\newcommand{\rad}{\operatorname{rad}}
\newcommand{\gen}{\operatorname{span}}
\noindent{\bf\red Proof of Theorem 1.}
{\blue I'm using $A$ and $B$ both as pushforwards and subspaces for now, can change later. Also need to define direct sum of PQs. I'm also using $\ann_Q(S)$ to mean $\ann_Q(S \cap \calD_Q)$ to not have to write $\calD_Q$ all the time.}
Uniqueness: If $A$ and $B$ are 2 pushforwards, then $\sigma_W(U+A) = \sigma_W(U+B)$ for all PQs $U$ on $W$.
Thus $\calD_A = \calD_B$, because otherwise if $w \in \calD_A\setminus \calD_B$, by taking $U(w) = 1$ on $\calD_U = \gen\{w\}$, we get $\sigma_W(U+A) = 1 \neq 0 = \sigma_W(U+B)$. Furthermore, $A$ and $B$ must agree where they are both defined, because by taking $U(w) = \frac{-A(w)-B(w)}{2}$ on $\calD_U = \gen\{w\}$ we get $(U+A)(w) = \frac{A(w)-B(w)}{2} = -(U+B)(w)$, so we must have $A(w) = B(w)$ to satisfy $\sigma_W(U+A) = \sigma_W(U+B)$. \\
Existence: Define $\phi_*Q$ by $\calD_{\phi_*Q} =\phi (\operatorname{ann}_Q(\ker\phi))$ and $\phi_*Q(w) = Q(v)$ where $v \in \ann_Q(\ker\phi)$. Note that $\phi_*Q$ is well-defined.
First consider when $U=0$ on all of $W$. Let $K$ be a maximal non-degenerate subspace of $\ker\phi$. Then $Q = Q\vert_K \oplus Q\vert_{\ann_Q(K)}$, and we can write $\ann_Q(K) = R \oplus A \oplus B$ where $R = \rad_Q(\ker\phi)$ and $A, B$ are chosen so that $A \subseteq \ann_Q(R)$ and $B \subseteq \ann_Q(K) \setminus \ann_Q(R)$. Since $Q: R \to B^*$ is surjective, for any $v \in \calD_Q$ there is some $r_v \in R$ such that $Q(r_v, B) = Q(v, B)$. If we choose the $r_v$ so that $r_{v_1}+r_{v_2} = r_{v_1+v_2}$, then we can replace $A$ by $A' = \{a-r_a : a \in A\}$ and $B$ by $B' = \{b-\frac{1}{2}r_b : b \in B\}$ to get $Q = Q\vert_K \oplus Q \vert_{R \oplus B'} \oplus Q\vert_{A'}$. Then notice that\begin{itemize} \item $\sigma_V(Q\vert_K) = \sigma_{\ker \phi}(Q\vert_{\ker\phi})$
\item $\sigma_V(Q \vert_{R \oplus B'}) = 0$
\item $\sigma_V(Q\vert_{A'}) = \sigma_W(\phi_*Q)$ \end{itemize}
so we get $\sigma_V(Q) = \sigma_{\ker \phi}(Q\vert_{\ker\phi}) + \sigma_W(\phi_*Q)$.
Now for an arbitrary $U$, note that $(Q + \phi^*U)\vert_{\ker \phi} = Q\vert_{\ker \phi}$ and $\phi_*(Q + \phi^*U) = \phi_*Q + U$ so we can replace $Q$ in the $U=0$ case by $Q + \phi^*U$ to get the general case.
\noindent{\bf\red Proof of Theorem 2.}