(Updated, as I realized the formula stated by Huybrechts indeed also does make sense without making any identifications.)

One should interpret the two sides in the compatibility condition of Huybrechts as defining two morphisms taking sections of $E$ to sections of $E \otimes \Omega^1$, i.e., if $\xi$ is a section of $E$ (over $V \subseteq U_i \cap U_j$), then the morphism on the left-hand side is $$\xi \mapsto\psi_i^{-1} ((d + A_i) (\psi_{i} \xi)),$$ which does indeed make sense as $\psi_i^{-1} \xi$ is a section of $V \times \mathbb{C}^r$. By a similar calculation as the one in Huybrechts after equation (4.4), one may show that this condition is equivalent to the compatibility condition that you write. I suppose Huybrechts has written it the way he did since he had not stated the compatibility condition that you write earlier in the text.

If one starts with your equation for compatibility, then one may compose this with $\psi_j^{-1}$ from the left and $\psi_j$ from the right, which yields $$ \psi_j^{-1} \psi_{ij}^{-1} d\psi_{ij} \psi_j = \psi_j^{-1} A_j \psi_j - \psi_i^{-1} A_i \psi_i $$ and then conclude the proof as in the last paragraph of Proposition 4.2.19. This would make the proof a bit shorter, but relies on the fact that one knows this equation for compatibility.

(Updated, as I realized the formula stated by Huybrechts indeed also does make sense without making any identifications.)

One should interpret the two sides in the compatibility condition of Huybrechts as defining two morphisms taking sections of $E$ to sections of $E \otimes \Omega^1$, i.e., if $\xi$ is a section of $E$ (over $V \subseteq U_i \cap U_j$), then the morphism on the left-hand side is $$\xi \mapsto\psi_i^{-1} ((d + A_i) (\psi_{i} \xi)),$$ which does indeed make sense as $\psi_i \xi$ is a section of $V \times \mathbb{C}^r$. By a similar calculation as the one in Huybrechts after equation (4.4), one may show that this condition is equivalent to the compatibility condition that you write. I suppose Huybrechts has written it the way he did since he had not stated the compatibility condition that you write earlier in the text.

If one starts with your equation for compatibility, then one may compose this with $\psi_j^{-1}$ from the left and $\psi_j$ from the right, which yields $$ \psi_j^{-1} \psi_{ij}^{-1} d\psi_{ij} \psi_j = \psi_j^{-1} A_j \psi_j - \psi_i^{-1} A_i \psi_i $$ and then conclude the proof as in the last paragraph of Proposition 4.2.19. This would make the proof a bit shorter, but relies on the fact that one knows this equation for compatibility.

I agree that the claimed formula does not seem to make sense without using some identification of $\psi_j$ and $\psi_i$ as matrices. If one chooses some fixed local trivialization of $E$, then one might of course do so, and then what is written in the proof after equation (4.4) seems to be a derivation of how the two gluing conditions are equivalent (and thus the first one is independent of the choice of fixed trivialization).

To prove the proposition, I would suggest instead to start with your equation for the gluing, compose with $\psi_j^{-1}$ from the left and $\psi_j$ from the right, which yields $$ \psi_j^{-1} \psi_{ij}^{-1} d\psi_{ij} \psi_j = \psi_j^{-1} A_j \psi_j - \psi_i^{-1} A_i \psi_i $$ and then conclude the proof as in the last paragraph of Proposition 4.2.19.

(Updated, as I realized the formula stated by Huybrechts indeed also does make sense without making any identifications.)

One should interpret the two sides in the compatibility condition of Huybrechts as defining two morphisms taking sections of $E$ to sections of $E \otimes \Omega^1$, i.e., if $\xi$ is a section of $E$ (over $V \subseteq U_i \cap U_j$), then the morphism on the left-hand side is $$\xi \mapsto\psi_i^{-1} ((d + A_i) (\psi_{i} \xi)),$$ which does indeed make sense as $\psi_i^{-1} \xi$ is a section of $V \times \mathbb{C}^r$. By a similar calculation as the one in Huybrechts after equation (4.4), one may show that this condition is equivalent to the compatibility condition that you write. I suppose Huybrechts has written it the way he did since he had not stated the compatibility condition that you write earlier in the text.

If one starts with your equation for compatibility, then one may compose this with $\psi_j^{-1}$ from the left and $\psi_j$ from the right, which yields $$ \psi_j^{-1} \psi_{ij}^{-1} d\psi_{ij} \psi_j = \psi_j^{-1} A_j \psi_j - \psi_i^{-1} A_i \psi_i $$ and then conclude the proof as in the last paragraph of Proposition 4.2.19. This would make the proof a bit shorter, but relies on the fact that one knows this equation for compatibility.