The tangent map of the multiplication map of a vector bundle If $\beta: \mathbf{U}\times \mathbf{V}\to \mathbf{W}$ is a bilinear map between real linear spaces then its derivative at a point $(u,v)$ is given by the Leibniz rule $$D\beta(u,v)(u_0,v_0)=\beta(u,v_0)+\beta(u_0,v).$$
Consequently the derivative of the multiplication map of a real linear space $\mu:\mathbb R\times \mathbf{V}\to \mathbf{V}$ is dual number multiplication $D\mu (\lambda,x)(\alpha,v)=\lambda v+\alpha x$.
Now suppose $\pi:E\to Y$ is a vector bundle, with multiplication map $\mu:\mathbb R\times E\to E$ over $Y$. Is it possible to calculate the tangent map $\mathrm T \mu: \mathrm T(\mathbb R \times E)\to \mathrm TE$, even though the multiplication map is only fiberwise bilinear?
Remark. My question is motivated by this one, whose answer I do not understand.
 A: Go back to the definition. Let $\gamma$ be a curve in $\mathbb{R}\times E$, we can take local coordinates on $E$ using the local trivialization, so
$$ \mathbb{R} \times E \ni (\lambda(s), v(s), b(s)) = \gamma(s) $$
The multiplication map gives
$$ \mu(\gamma(s)) = (\lambda(s) v(s), b(s)) $$
From this you find immediately that the derivative at $0$ is
$$   (\dot{\lambda}(0) v(0) + \lambda(0) \dot{v}(0), \dot{b}(0) ) \tag1$$

More generally: the tangent space to a Cartesian product $A\times B$ at a point $(a,b)$ is isomorphic to $T_a A \oplus T_b B$.
Now, take $\varphi:A\times B \to C$. Fix $a\in A$ and $b\in B$. Let $\sigma(\bullet) = \varphi(\bullet, b)$ and $\tau(\bullet) = \varphi(a,\bullet)$. Then
$$ T_{(a,b)}\varphi(\alpha,\beta) = T_a\sigma(\alpha) + T_b\tau(\beta) $$
exactly like in multivariable calculus.
In your case, with $\varphi$ the multiplication operator, you have that $\tau$ is the smooth map $E\to E$ given by multiplication by the scalar $a$. And $\sigma$ is the the map $\mathbb{R}\to E$ given by multiplying the fibre of a fixed element $b\in E$.
Perhaps what is confusing you is the fact that given a vector bundle $E \to Y$ and $e\in E$, that there is (via the linear structure on the fibre) a canonical identification of the fibre $\pi^{-1}(\pi(e))$ over $\pi(e)$ with the vertical part $V_eE$ of the tangent space (the kernel of $T_e\pi$ as a mapping from $T_e E \to T_{\pi(e)} Y$). If you denote by $\psi_e: \pi^{-1}(\pi(e)) \to V_e E$ this mapping, then you can write (1) as
$$ T_{(\lambda,e)}\mu(\alpha,v) = \lambda v + \alpha \psi_e(e) $$
