Let $A$ be an $n \times m$ matrix, $Y$ be an $m \times k$ matrix, and $B$ be an $n \times k$ matrix such that
$$Y = A \ B = A^+ \ B,$$
where $A^+$ is a Moore-Penrose pseudoinverse, given by
$$
A^+ = \begin{cases}
A^\top (AA^\top)^{-1} & n \le m\
(A^\top A)^{-1} A^\top & n \ge m\
\end{cases}.
$$
Let $C = A^+$.
Then
$$
\dot{C} = \begin{cases}
\dot{A}^\top (AA^\top)^{-1} - A^\top (AA^\top)^{-1} (\dot{A} A^\top + A \dot{A}^\top) (AA^\top)^{-1} & n \le m\\
(A^\top A)^{-1} \dot{A}^\top -(A^\top A)^{-1} (\dot{A}^\top A + A^\top \dot{A}) (A^\top A)^{-1} A^\top & n \ge m\\
\end{cases}\\
= \begin{cases}
(I - CA) \dot{A}^\top (AA^\top)^{-1} - C \dot{A} C & n \le m\\
(A^\top A)^{-1} \dot{A}^\top (I - AC) - C \dot{A}C & n \ge m\\
\end{cases}\\
= \begin{cases}
(I - CA) \dot{A}^\top C^\top C - C \dot{A} C & n \le m\\
CC^\top \dot{A}^\top (I - AC) - C \dot{A}C & n \ge m\\
\end{cases}\\
= (I - CA) \dot{A}^\top C^\top C + CC^\top \dot{A}^\top (I - AC) - C \dot{A} C.
$$
Then, we can work out the pullback:
$$
\langle \overline{C}, \dot{C} \rangle =
\langle\overline{C}, - C \dot{A} C\rangle + \langle\overline{C},(I - CA) \dot{A}^\top C^\top C\rangle + \langle\overline{C},C C^\top \dot{A}^\top (I - AC)\rangle\
= \langle-C^\top\overline{C}C^\top + C^\top C\overline{C}^\top (I - CA) + (I - AC)\overline{C}^\top C C^\top, \dot{A} \rangle\
\overline{A} = -C^\top\overline{C}C^\top + C^\top C\overline{C}^\top (I - CA) + (I - AC)\overline{C}^\top C C^\top
$$
For the next step, we have $Y = C B$, so $\dot{Y} = \dot{C}B + C \dot{B}$ and
$$
\overline{B} = C^\top\overline{Y}\\
\overline{C} = \overline{Y}B^\top
$$
Combining, we have
$$
\overline{A} = -C^\top\overline{Y}B^\top C^\top + C^\top CB\overline{Y}^\top (I - CA) + (I - AC)B\overline{Y}^\top C C^\top\
=-\overline{B}Y^\top + C^\top Y (\overline{Y}^\top - \overline{B}^\top A) + (B - AY)\overline{B}^\top C^\top\
=-\overline{B}Y^\top + A^\top \backslash Y (\overline{Y}^\top - \overline{B}^\top A) + (B - AY)\overline{B}^\top / A^\top.
$$
