If $E=F\oplus G$ and $u\in\Hom(E)$ satisfies: $u(F)\sbe F$ and $u(G)\sbe G$, then $\det u=\det(u|F)\det(u|G)$.
Let $f_1,\ldots,f_k$ and $g_1,\ldots,g_m$ be bases for $F$ and $G$ respectively and choose a volume form on $E$ such that $\vol{}(f_1,\ldots,g_m)=1$, then
$$
\o:(x_1,\ldots,x_k)\mapsto\vol{}(x_1,\ldots,x_k,g_1,\ldots,g_m)
$$
is a volume form on $F$ and therefore:
\begin{eqnarray*}
\vol{}(u(f_1),\ldots,u(f_k),g_1,\ldots,g_m)
&=&\o(u(f_1),\ldots,u(f_k))\\
&=&\det(u|F)\o(f_1,\ldots,f_k)\\
&=&\det(u|F)\vol{}(f_1,\ldots,g_m)
=\det(u|F)~.
\end{eqnarray*}
Now $\eta:(y_1,\ldots,y_m)\mapsto\vol{}(u(f_1),\ldots,u(f_k),y_1,\ldots,y_m)$ is a volume form on $G$, which implies:
\begin{eqnarray*}
\vol{}(u(f_1),\ldots,u(f_k),u(g_1),\ldots,u(g_m))
&=&\eta(u(g_1),\ldots,u(g_m))\\
&=&\det(u|G)\eta(g_1,\ldots,g_m)\\
&=&\det(u|G)\vol{}(u(f_1),\ldots,u(f_k),g_1,\ldots,g_m)
=\det(u|G)\det(u|F)~.
\end{eqnarray*}
On the other hand the left hand side equals by definition $\det(u)$.