[proofplan]
The determinant line bundle is the top exterior power $\Lambda^r E$, so each local trivialization of $E$ induces a local trivialization of $\Lambda^r E$. On overlaps, the induced transition map is the top exterior power of the original transition map $g_{ij}(x):\mathbb R^r\to\mathbb R^r$. Since $\Lambda^r\mathbb R^r$ is one-dimensional, the [linear map](/page/Linear%20Map) $\Lambda^r g_{ij}(x)$ is multiplication by $\det(g_{ij}(x))$, giving exactly the claimed scalar transition functions.
[/proofplan]
[step:Construct the determinant trivializations from the vector bundle trivializations]
Let $e_1,\dots,e_r\in\mathbb R^r$ denote the standard basis vectors, and define
\begin{align*}
\varepsilon:=e_1\wedge\cdots\wedge e_r\in\Lambda^r\mathbb R^r.
\end{align*}
For each $i\in I$, the trivialization $\varphi_i$ induces a smooth vector bundle trivialization
\begin{align*}
\Lambda^r\varphi_i:(\Lambda^r E)|_{U_i}&\to U_i\times\Lambda^r\mathbb R^r
\end{align*}
defined fiberwise as follows: if $x\in U_i$ and $\xi_1,\dots,\xi_r\in E_x$, then
\begin{align*}
\Lambda^r\varphi_i(\xi_1\wedge\cdots\wedge\xi_r)
=
\left(x,\operatorname{pr}_2(\varphi_i(\xi_1))\wedge\cdots\wedge\operatorname{pr}_2(\varphi_i(\xi_r))\right),
\end{align*}
where $\operatorname{pr}_2:U_i\times\mathbb R^r\to\mathbb R^r$ is the projection onto the second factor.
Define the linear isomorphism
\begin{align*}
\theta:\Lambda^r\mathbb R^r&\to\mathbb R
\end{align*}
by $\theta(\varepsilon)=1$. Composing $\Lambda^r\varphi_i$ with $\operatorname{id}_{U_i}\times\theta$ gives a line-bundle trivialization
\begin{align*}
\psi_i:(\Lambda^r E)|_{U_i}&\to U_i\times\mathbb R.
\end{align*}
These are the local trivializations of $\det E=\Lambda^r E$ used to compute the transition functions.
[/step]
[step:Compute the induced transition map on the top exterior power]
Fix $i,j\in I$ and let $x\in U_i\cap U_j$. Define the fiber transition map
\begin{align*}
G_{ij}(x):\mathbb R^r&\to\mathbb R^r
\end{align*}
by $G_{ij}(x)(v)=g_{ij}(x)v$. The transition relation for $E$ says that, for every $v\in\mathbb R^r$,
\begin{align*}
\varphi_i\circ\varphi_j^{-1}(x,v)=(x,G_{ij}(x)(v)).
\end{align*}
Taking the $r$-th exterior power fiberwise, the corresponding transition map for $\Lambda^r E$ before identifying $\Lambda^r\mathbb R^r$ with $\mathbb R$ is
\begin{align*}
\Lambda^r G_{ij}(x):\Lambda^r\mathbb R^r&\to\Lambda^r\mathbb R^r.
\end{align*}
[guided]
Fix $i,j\in I$ and a point $x\in U_i\cap U_j$. The transition function $g_{ij}$ assigns to $x$ an invertible matrix $g_{ij}(x)\in GL(r,\mathbb R)$. We regard this matrix as the linear map
\begin{align*}
G_{ij}(x):\mathbb R^r&\to\mathbb R^r
\end{align*}
defined by $G_{ij}(x)(v)=g_{ij}(x)v$.
The original transition identity for $E$ is
\begin{align*}
\varphi_i\circ\varphi_j^{-1}(x,v)=(x,G_{ij}(x)(v)).
\end{align*}
This says that changing from the $j$-trivialization to the $i$-trivialization applies the linear map $G_{ij}(x)$ to the fiber coordinate.
Now pass to the determinant line $\Lambda^r E$. A decomposable element of the fiber $\Lambda^r E_x$ has the form $\xi_1\wedge\cdots\wedge\xi_r$ with $\xi_1,\dots,\xi_r\in E_x$. Applying the induced trivializations sends each $\xi_k$ to its coordinate vector in $\mathbb R^r$, and therefore the overlap map applies $G_{ij}(x)$ to each of the $r$ coordinate vectors. Thus the induced fiber map is
\begin{align*}
\Lambda^r G_{ij}(x):\Lambda^r\mathbb R^r&\to\Lambda^r\mathbb R^r,
\end{align*}
where
\begin{align*}
(\Lambda^r G_{ij}(x))(v_1\wedge\cdots\wedge v_r)
=
G_{ij}(x)(v_1)\wedge\cdots\wedge G_{ij}(x)(v_r)
\end{align*}
for all $v_1,\dots,v_r\in\mathbb R^r$. This is exactly the exterior-power construction applied to the fiber transition map.
[/guided]
[/step]
[step:Identify the top exterior power action with multiplication by the determinant]
Let $A:\mathbb R^r\to\mathbb R^r$ be a linear isomorphism, and let $J_A\in\mathbb R^{r\times r}$ be its matrix in the standard basis $e_1,\dots,e_r$. By the defining alternating multilinear formula for the determinant,
\begin{align*}
(\Lambda^r A)(e_1\wedge\cdots\wedge e_r)
=
A e_1\wedge\cdots\wedge A e_r
=
\det(J_A)(e_1\wedge\cdots\wedge e_r).
\end{align*}
Since $\Lambda^r\mathbb R^r$ is spanned by $\varepsilon=e_1\wedge\cdots\wedge e_r$, this proves that $\Lambda^r A$ is multiplication by $\det(J_A)$.
Apply this with $A=G_{ij}(x)$. The matrix of $G_{ij}(x)$ in the standard basis is $g_{ij}(x)$, so
\begin{align*}
(\Lambda^r G_{ij}(x))(\varepsilon)=\det(g_{ij}(x))\varepsilon.
\end{align*}
After applying $\theta:\Lambda^r\mathbb R^r\to\mathbb R$, the scalar transition map is therefore
\begin{align*}
x\mapsto \det(g_{ij}(x)).
\end{align*}
[/step]
[step:Verify smoothness and the cocycle identity for the determinant transitions]
Because $g_{ij}:U_i\cap U_j\to GL(r,\mathbb R)$ is smooth and the determinant map
\begin{align*}
\det:GL(r,\mathbb R)&\to\mathbb R^\times
\end{align*}
is a polynomial function in the matrix entries restricted to $GL(r,\mathbb R)$, the composition
\begin{align*}
\det(g_{ij}):U_i\cap U_j&\to\mathbb R^\times
\end{align*}
is smooth.
On a triple overlap $U_i\cap U_j\cap U_k$, the transition functions for $E$ satisfy
\begin{align*}
g_{ik}(x)=g_{ij}(x)g_{jk}(x)
\end{align*}
for every $x\in U_i\cap U_j\cap U_k$. Taking determinants and using multiplicativity of the determinant gives
\begin{align*}
\det(g_{ik}(x))
=
\det(g_{ij}(x)g_{jk}(x))
=
\det(g_{ij}(x))\det(g_{jk}(x)).
\end{align*}
Thus the maps $\det(g_{ij})$ satisfy the line-bundle cocycle identity. Hence, relative to the determinant trivializations $\psi_i$, the determinant line bundle $\det E=\Lambda^r E$ has transition functions $\det(g_{ij}):U_i\cap U_j\to\mathbb R^\times$, as claimed.
[/step]
