Assume $f:E\rightarrow X\times [a,b]$ is a fiber bundle. Assume $X\times [a,c]$ and $X\times [c,b]$ for some $a<c<b$ are both trivial bundles. Let $h_1:f^{-1}(X\times [a,c])\rightarrow X\times [a,c]\times F$ and $h_1:f^{-1}(X\times [c,b])\rightarrow X\times [c,b]\times F$ be fiber bundles isomorphisms. These maps do not necessarily agree on $f^{-1}(X\times c)$ so we make them agree by 'replacing $h_2$ with the isomorphism $X\times[c,b]\times F\rightarrow X\times [c,b]\times F$' that is given on each slice $X\times x\times F$ by the map $h_1h_2^{-1}:X\times c\times F\rightarrow X\times c\times F$
This is essentially the argument in https://pi.math.cornell.edu/~hatcher/VBKT/VB.pdf (page 21). I don't understand how to explicitly define the isomorphism and why it is even well defined.
Could someone elaborate? without going into things like $Gl(n,K)$ and so on please :)
Write $E_1 = f^{-1}(X\times[a,c])$ and $E_2 = f^{-1}(X \times [c, b])$. By assumption, we have trivializations on $E_1$ and $E_2$; let's call them $h_1 \colon E_1 \to X \times [a,c] \times F$ and $h_2 \colon E_2 \to X \times [c, b] \times F$.
Here's some notation to make everything super explicit:
Let $\pi_1\colon X \times [c, b] \times F \to X \times F$ be the map defined by $\pi_1(x, t, v) = (x, v)$.
Let $\pi_2 \colon X \times [c, b] \times F \to [c, b]$ be the map defined by $\pi_1(x, t, v) = t$.
Define $i_c\colon X\times F \to X\times \{c\} \times F$ by $i_c(x, v)= (x,c,v)$. Note that $i_c\pi_1 = \operatorname{id}_{X\times \{c\} \times F}$.
Define $\phi \colon X \times F \to X \times F$ by $\phi(x ,v) = \pi_1 h_1 h_2^{-1}(x, c, v)$, so $\phi = \pi_1h_1h_2^{-1}i_c$
We define a map $\alpha \colon X \times [c, b] \times F \to X \times [c, b] \times F$ by putting $$ \pi_1(\alpha(x, t, v)) = \phi(x, v) $$ and $$ \pi_2(\alpha(x,t,v)) = t. $$ These defining properties $\pi_1 \circ \alpha = \phi \circ \pi_1$ and $\pi_2 \circ \alpha = \pi_2$ specify $\alpha$ using the universal property of the product.
Now define $h\colon E \to X \times [a, b] \times F$ by $$ h(e) = \begin{cases} h_1(e) & e \in E_1 \\ \alpha(h_2(e)) & e \in E_2 \\ \end{cases} $$ This is well-defined because if $e \in E_1 \cap E_2$ then $$ \pi_1(\alpha(h_2(e))) = \phi\pi_1 h_2 e = \pi_1h_1h_2^{-1}i_c \pi_1 h_2 e = \pi_1h_1h_2^{-1} h_2e = \pi_1(h_1(e)) $$ and $$ \pi_2(\alpha(h_2(e)) = \pi_2(h_2(e)), $$ so $h_1(e) = \alpha(h_2(e))$.