I'm currently reading Riemannian Geometry by Saski, and just finished reading Proposition 4.1 regarding the connection map of a an affine connection. In particular, suppose that $M$ is a smooth manifold. Then the vertical subspace at $(p, v) \in TM$ is defined as $V_{(p, v)} = T_{(p,v)} T_p M$ (or equivalently, as the kernel of $\tau_{M\ast}$, where $\tau_M: TM \to M$ is the canonical projection).
The connection map $K_{(p,v)}: T_{(p,v)}TM \to T_pM$ is then defined in coordinates on $TM$ by $$K_{(p,v)}\left(\alpha^i \frac{\partial}{\partial q^i} \Big{\vert}_{(p,v)} + \beta^i \frac{\partial}{\partial \xi^i} \Big{\vert}_{(p,v)}\right) = \left(\beta^i + \alpha^j v^k \Gamma^i_{jk}\right) \frac{\partial}{\partial q^i} \Big{\vert}_{p}$$
Where $\Gamma_{jk}^i$ are the Christoffel symbols of the connection $\nabla$, and $v = v^i \frac{\partial}{\partial q^i}$ in coordinates on $M$. Moreover, we have the induced map $K: TTM \to TM$ which agrees with $K_{(p,v)}$ on each tangent space.
The kernel of $K_{(p,v)}$ then defines the horizontal subspace $H_{(p,v)}$ in the decomposition $$T_{(p,v)}TM = H_{(p,v)} \oplus V_{(p,v)}$$ for all $(p, v)\in TM$.
It is easy to see that if $X\in \mathcal{X}(M)$, then the vector $\tilde{X} \in T_{(p, X(p))}TM$ given by $$\tilde{X}_v = v^i \frac{\partial}{\partial q^i} \Big{\vert}_{(p,X(p))} + v^j \frac{\partial X^i}{\partial q^j} \frac{\partial}{\partial \xi^i} \Big{\vert}_{(p,X(p))}$$ satisfies $\nabla_v X = K(\tilde{X}_v)$.
Following this, Exercise 1 claims that given a map $K: TTQ \to TQ$ such that the following diagrams commute, and $K$ is a vector bundle map for each one:
and such that $$K\left( \beta^i \frac{\partial}{\partial \xi^i} \Big{\vert}_{(p,v)} \right) = \beta^i \frac{\partial}{\partial q^i} \Big{\vert}_{p}$$ for all $\beta^i$, then the map $\nabla_v X := K(\tilde{X}_v)$ is an affine connection.
It's clear to me that such a $\nabla$ must be $\mathbb{R}$-linear in both arguments, but I'm having trouble seeing that $$\nabla_v (fX) = v(f)X + f \nabla_v X$$
By direct calculation, $$K(\widetilde{fX}_v) = v^i K\left(\frac{\partial}{\partial q^i} \Big{\vert}_{(p,f(p)X(p))}\right) + v(f)X(p) + f(p) v^j \frac{\partial X^i}{\partial q^j} \frac{\partial}{\partial q^i} \Big{\vert}_{p}$$ and $$f(p)K(\tilde{X}_v) = f(p) v^i K\left(\frac{\partial}{\partial q^i} \Big{\vert}_{(p,X(p))}\right) + f(p) v^j \frac{\partial X^i}{\partial q^j} \frac{\partial}{\partial q^i} \Big{\vert}_{p}$$
Why should I expect that $f(p)K\left(\frac{\partial}{\partial q^i} \Big{\vert}_{(p,X(p))}\right) = K\left(\frac{\partial}{\partial q^i} \Big{\vert}_{(p,f(p)X(p))}\right)$?