What’s in the link is actually not a more general definition, it is more restrictive. Here’s what Hamilton writes; immediately after giving the definition of a principal bundle, he says
Remark 4.1.3: The classic references [133] (Steenrod) and [81] (Husemoller) use the term fibre bundle in a
more restrictive sense; see Remark 4.1.15.
and remark 4.1.15 says the following:
Remark 4.1.15: Some references, such as [133] and [81], use the term fibre bundle more restrictively. If the topological definition in these books is transferred to a
smooth setting, the definition amounts to assuming that the transition functions of a
bundle atlas are smooth maps to a Lie group $G$, acting smoothly as a transformation
group on the fibre $F$, instead of maps to the full diffeomorphism group $\text{Diff}(F)$ of
the fibre:
\begin{align}
\phi_{ji}:U_i\cap U_j&\to G\\
x&\mapsto \phi_{jx}\circ\phi_{ix}^{-1}.
\end{align}
Equivalently, a fibre bundle is with this definition always an associated bundle in the
sense of Remark 4.7.8.
So, fiber bundle as defined by Hamilton (and also other authors) does not come equipped with a group, whereas in the other references of Steenrod and Husemoller (let’s just take Husemoller for example) the definition is as the associated fiber bundle:
Definition.[Husemoller, Definition of Fiber Bundle]
Let $\xi=(X,p,B)$ be a principal $G$-bundle and let $F$ be a left $G$-space. The relation $(x,y)s=(xs,s^{-1}y)$ defines a right $G$-space structure on $X\times F$. Let $X_F$ denote the quotient space $(X\times F)/G$ and $p_F$ the factorization of $X\times F\to X\to B$ by the projection $X\times F\to X_F$. With this notation, the bundle $(X_F,p_F,B)$, denoted $\xi[F]$, is called the fiber bundle over $B$ with fiber $F$ (viewed as a $G$-space) and associated principal bundle $\xi$. The group $G$ is called the structure group of the fiber bundle $\xi[F]$.
Reiterating, note that although in this source one calls $G$ “the structure group of $\xi[F]$”, the very definition of fiber bundle is restrictive, in the sense that a fiber bundle is defined to be an associated bundle of a principal bundle (whereas Hamilton requires it to be a triple $(X,p,B)$ where $p$ is surjective and smooth/continuous and satisfy a local-triviality condition… other names for this object include locally-trivial (smooth/continuous) bundle).
Best Answer
It's not trivial in general. For example, the Klein bottle $K$ is the total space of a fibre bundle $S^1 \to K \xrightarrow{\pi} S^1$. Note that $\pi^*TS^1$ is trivial and if $V$ were isomorphic to $TS^1$, it would also be trivial. This would imply that $TK$ is trivial, but this is not the case as $K$ is non-orientable.
Let $i : F \to E$ be the inclusion of a fiber. Since $\pi\circ i$ is constant, we see that $0 = (\pi\circ i)_* = \pi_*\circ i_*$, so $i_*TF \subseteq \ker \pi_* = V$. By comparing ranks, we see that in fact $i_*TF = V$ and hence $TF \cong i^*V$.