Curvature $\Omega$ is a $\mathfrak{g}$ valued $2$-form on $P$ i.e., for each $p\in P$, we have $\Omega(p):T_pP\times T_pP\rightarrow \mathfrak{g}$.

As $\mathfrak{g}$ is $\mathfrak{gl}(r,\mathbb{C})$, given $(v_1,v_2)\in T_pP\times T_pP$, we get a $r\times r$ matrix $\Omega(p)(v_1,v_2)=[a_{ij}]$.

Once we fix $(v_1,v_2)$ we get $a_{ij}\in \mathbb{C}$, these depend on $(v_1,v_2)$. So, we have $$\Omega(p)(v_1,v_2)=[a_{ij}(v_1,v_2)]$$ Here $a_{ij}$ are functions $a_{ij}:T_pP\times T_pP\rightarrow \mathbb{C}$ with $(v_1,v_2)\mapsto a_{ij}(v_1,v_2)\in \mathbb{C}$.

We can write $\Omega(p)(v_1,v_2)=[a_{ij}(v_1,v_2)]$ as $\Omega(p)=[a_{ij}]$.

So, for each $p\in P$, $\Omega(p)$ is a matrix of functions $a_{ij}:T_pP\times T_pP\rightarrow \mathbb{C}$.

Thus, we can write $\Omega$ as an $r\times r $ matrix $[\Omega_{ij}]$ where $\Omega_{ij}$ is a $\mathbb{C}$ valued $2$ form on $P$ given by $\Omega_{ij}(p):T_pP\times T_pP\rightarrow\mathbb{C}$ is the map $a_{ij}:T_pP\times T_pP\rightarrow \mathbb{C}$.

Thus, $I_r-\frac{1}{2\pi\sqrt{-1}}\Omega$ is an $r\times r$ matrix whose $i,j$-th entry is $\delta_{ij}-\frac{1}{2\pi \sqrt{-1}}\Omega_{ij}$.

Let $\omega,\tau$ be $p$-form and $q$-form respectively. Then, $\omega\wedge \tau=(-1)^{pq}\tau\wedge \omega$. We are only dealing with $2$-forms here. So, $pq=4$ and $\omega\wedge \tau=\tau\wedge \omega$ for each $\omega,\tau$.

We have a matrix $I_r-\frac{1}{2\pi\sqrt{-1}}\Omega$ whose entries are coming from a commutative ring (of $2$-forms).

We can then talk about determinant of that matrix and $\text{det}(I_r-\frac{1}{2\pi\sqrt{-1}}\Omega)$ is precisely that.

Curvature $\Omega$ is a $\mathfrak{g}$ valued $2$-form on $P$ i.e., for each $p\in P$, we have $\Omega(p):T_pP\times T_pP\rightarrow \mathfrak{g}$.

As $\mathfrak{g}$ is $\mathfrak{gl}(r,\mathbb{C})$, given $(v_1,v_2)\in T_pP\times T_pP$, we get a $r\times r$ matrix $\Omega(p)(v_1,v_2)=[a_{ij}]$.

Once we fix $(v_1,v_2)$ we get $a_{ij}\in \mathbb{C}$, these depend on $(v_1,v_2)$. So, we have $$\Omega(p)(v_1,v_2)=[a_{ij}(v_1,v_2)]$$ Here $a_{ij}$ are functions $a_{ij}:T_pP\times T_pP\rightarrow \mathbb{C}$ with $(v_1,v_2)\mapsto a_{ij}(v_1,v_2)\in \mathbb{C}$.

We can write $\Omega(p)(v_1,v_2)=[a_{ij}(v_1,v_2)]$ as $\Omega(p)=[a_{ij}]$.

So, for each $p\in P$, $\Omega(p)$ is a matrix of functions $a_{ij}:T_pP\times T_pP\rightarrow \mathbb{C}$.

Thus, we can write $\Omega$ as an $r\times r $ matrix $[\Omega_{ij}]$ where $\Omega_{ij}$ is a $\mathbb{C}$ valued $2$ form on $P$ given by $\Omega_{ij}(p):T_pP\times T_pP\rightarrow\mathbb{C}$ is the map $a_{ij}:T_pP\times T_pP\rightarrow \mathbb{C}$.

Thus, $I_r-\frac{1}{2\pi\sqrt{-1}}\Omega$ is an $r\times r$ matrix whose $i,j$-th entry is $\delta_{ij}-\frac{1}{2\pi \sqrt{-1}}\Omega_{ij}$.

Let $\omega,\tau$ be $p$-form and $q$-form respectively. Then, $\omega\wedge \tau=(-1)^{pq}\tau\wedge \omega$. We are only dealing with $2$-forms here. So, $pq=4$ and $\omega\wedge \tau=\tau\wedge \omega$ for each $\omega,\tau$.

We have a matrix $I_r-\frac{1}{2\pi\sqrt{-1}}\Omega$ whose entries are coming from a commutative ring (of $2$-forms).

We can then talk about determinant of that matrix and $\text{det}(I_r-\frac{1}{2\pi\sqrt{-1}}\Omega)$ is precisely that.

For simplicity, take $r=2$. Then, $$\text{det}(I-\Omega)=\text{det}\begin{bmatrix}1-\Omega_{11}&-\Omega_{12}\\ -\Omega_{21}&1-\Omega_{22}\end{bmatrix}=1-\underbrace{(\Omega_{11}+\Omega_{22})}_{2-form}+\underbrace{\Omega_{11}\wedge\Omega_{12}-\Omega_{21}\wedge \Omega_{12}}_{4-form}$$ We then have $$p^*(1+\gamma_1+\gamma_2)=1+\underbrace{p^*(\gamma_1)}_{2-form}++\underbrace{p^*(\gamma_2)}_{4-form}$$ So, it makes sense to say $\text{det}(I-\Omega)=p^*(1+\gamma_1+\gamma_2)$ (I ignored $\frac{1}{2\pi \sqrt{-1}}$ to keep it simple).

We have $p^*(\gamma_1)=-\Omega_{11}-\Omega_{22}$ and $p^*(\gamma_2)=\Omega_{11}\wedge\Omega_{12}-\Omega_{21}\wedge \Omega_{12}$