I want to show that there is a canonical isomorphism $ T_b B \oplus E_b\cong T_{(b,0)}E$.
If this map is to be canonical some natural choices are to take the the derivatives of the inclusions $i:E_b\rightarrow E$, and $j:B\rightarrow E$.
Then we have a map $F= di(b)+dj(0): T_b B \oplus E_b\rightarrow T_{(b,0)}E$ but it's not obvious that this is an isomorphism.
I know we could also attack this problem by showing there is a split exact sequence, but I'm not sure if the split exact sequences gives that the isomorphism is canonical.
Fix $b\in B$. The injection $i : E_b \hookrightarrow E$ and the projection $\pi : E \to B$ gives the exact sequence
$$0 \to E_b \to T_{(b,0)}E \to T_bB \to 0$$
where the first map is given by $d_0i$ (we identify $E_b\simeq T_0E_b$) and the second one by $d_b\pi$. The sequence is indeed exact: $d_b\pi \circ d_0 i = d_0(\pi\circ i)=0$ and $\pi \circ i$ is the constant map $b:E_b \to B$, hence $im(d_0i)\subseteq ker(d_b\pi)$. But these space have same dimension (rank of $E$) so they are equal.
Moreover, this sequence splits: take the zero section $j : B \to E$, then $d_bj : T_bB \to T_{(b,0)}E$ gives a section (in the set theoretic sense) of $d_b\pi : T_{(b,0)}E \to T_bB$ (indeed, $d_{(0,b)}\pi \circ d_bj = d_b(\pi \circ j) = d(id_{T_bB})=1$), so you obtain the isomorphism $$T_{(b,0)}E \simeq E_b\oplus T_bB.$$
This isomorphism is natural in the sense that the injection $i$ is natural after fixing $b\in B$.