I'll change your notation a little to make things clearer (in my opinion, at least). Let $\pi \colon E \rightarrow M$ be a smooth vector bundle. With it comes the associate short exact sequence
$$ 0 \rightarrow VE \hookrightarrow TE \xrightarrow{d\pi} \pi^{*}(TM) \rightarrow 0 $$
of vector bundles over $E$. For the purpose of defining the covariant derivative, it is better to consider a left splitting $K \colon TE \rightarrow VE$ (over $E$). Note that $VE \cong \pi^{*}(E)$ using the natural isomorphism which allows to identify $V_{(p,v)}E = T_{(p,v)}(E_p)$ (vectors which are tangent to the fiber $E_p$) with the vector space $E_p$. Denote this isomorphism by $\Phi$ and let $\pi_{\sharp} \colon \pi^{*}(E) \rightarrow E$ be the natural map of vector bundles that covers $\pi$. Then we can define the covariant derivative of a section $s \in \Gamma(E)$ by
$$ \nabla s = \pi_{\sharp} \circ \Phi \circ K \circ ds.$$
More explicitly, $s$ is a map from $M$ to $E$ and $ds \colon TM \rightarrow TE$ is the regular differential. To get the covariant derivative, we take the regular derivative $ds$, project it to the vertical space using $K$ and then identify the vertical space with $E$ to get back a section of $E$ over $M$. If the splitting $K$ satisfies the equivariance conditions appropriate for a connection on a vector bundle, this will reconstruct the usual covariant derivative.
Let us try and see concretely how the process above works when $E = M \times \mathbb{R}^k$ is the trivial bundle. Fix some coordinate neighborhood $U$ with coordinates $x^1,\dots,x^n$ and let $\xi^1,\dots,\xi^k$ denote the coordinates on $\mathbb{R}^k$. Then $\pi^{-1}(U)$ is a coordinate neighborhood with coordinates I'll denote by $\tilde{x}^1,\dots,\tilde{x}^n$ and $\tilde{\xi}^1,\dots,\tilde{\xi}^k$. We have $\tilde{x}^i = x^i \circ \pi_1$ and $\tilde{\xi}^i = \xi^i \circ \pi_2$ and I use the $\tilde \,$ to differentiate between the coordinates on the base / fiber and on the total space.
With this notation, the vertical space $V_{(p,v)}E$ at $(p,v)$ is precisely $$\operatorname{span} \left \{ \frac{\partial}{\partial \tilde{\xi}^1}|_{(p,v)}, \dots, \frac{\partial}{\partial \tilde{\xi}^k}|_{(p,v)} \right \}. $$
A projection $K$ from $TE$ onto $VE$ will look like:
$$ K|_{(p,v)} = a_i^j(p,v) d\tilde{x}^i \otimes \frac{\partial}{\partial \tilde{\xi}^j} + d\tilde{\xi}^i \otimes \frac{\partial}{\partial \tilde{\xi}^i}$$
(the image must be the vertical bundle and it must satisfy $K^2 = K$).
Now, let $s \colon M \rightarrow M \times \mathbb{R}^k$ be a section and write $s(p) = (p, f(p))$ for some $f = (f^1,\dots,f^k) \colon M \rightarrow \mathbb{R}^k$. Set
$$e_i(p) := (p, \underbrace{(0,\dots,0,1,0,\dots,0)}_{i\text{th place}}$$
to be the constant sections corresponding to the standard basis vectors so $s = f^i e_i$. Let us see how the covariant derivative of $s$ in the direction $\frac{\partial}{\partial x^l} = \partial_l$ (in the base) at the point $p$ looks like:
$$ ds|_{p} = dx^i \otimes \frac{\partial}{\partial \tilde{x}^i} + \frac{\partial f^i}{\partial x^j} dx^j \otimes \frac{\partial}{\partial \tilde{\xi}^i}, \\
K \circ ds
= \left( a_i^j(p,f(p)) + \frac{\partial f^j}{\partial x^i}(p) \right) dx^i \otimes \frac{\partial}{\partial \tilde{\xi}^j}, \\
\nabla_l(s)(p) = \left( a_l^j(p, f(p)) + \frac{\partial f^j}{\partial x^l}(p) \right) e_j(p). $$
Note that $\nabla_l(s)(p)$ has two components. The second is the regular directional derivative of the components of $s$ with respect to the frame $(e_1,\dots,e_k)$ in the direction $\partial_l$. The first comes from the the projection $K$. If $a_i^j \equiv 0$, this is gone. Also, the components $a_i^j$ depend both on the point $p$ and the value $f(p)$ (this reflects the fact that $K$ gives us a projection of $TE$ onto $\pi^{*}(E)$). For a general vector bundle, this is the local picture.
Regarding your questions, we're not ignoring the variation between fibers. This is encoded in the particular way $K$ projects onto $VE$ (through the coefficients $a_i^j$ which give rise under certain assumptions to the Christoffel symbols $\Gamma_{ik}^j$ of the connection). While the image of $K$ is always $VE$, the kernel of $KE$ is different at each point and provides us with the horizontal space. The horizontal space tells us how we should identify fibers infinitesimally along curves over the base space.
Covariant differentiation allows us to differentiate a section along a vector field on $M$ and get back a section. It is done by performing regular differentiation and obtaining a tangent vector in $E$ which is necessarily not tangent to the fiber. The connection mechanism, via $K$, provides us with a way to project this tangent vector in a consistent way to get a vector which is tangent to the fiber and then identify it with an element of the fiber.
Best Answer
This is just a slight expansion of Amitai Yuval's comment.
The problem is that integral curves are not defined with respect to a line subbundle of the tangent bundle, but with respect to a vector field. (Only the images of the integral curves are "known" to the horizontal subbundle, much like only integral submanifolds - without any parametrization - are "known" to general subbundles of the tangent bundle.) Geometrically this says the horizontal bundle does not determine how quickly to flow along its integral submanifolds (the images of the integral curves).
The horizontal subbundle amounts to a multiple of a global horizontal vector field by a scalar function. Of these it is naturally best to choose a lift of the global vector field $\partial/\partial t$ on $I$. With respect to this horizontal vector field the integral curves are now defined.
That this field lifts $\partial/\partial t$ implies by the chain rule that $\frac{\partial}{\partial t}(\pi\circ c_{(s,x)})\equiv 1$. But $\pi\circ c_{(s,x)}$ is precisely the map $f$ giving the first coordinate of the integral curve $c$. By assumption $f(0)=s$ and $f^\prime \equiv 1$. By uniqueness of solutions to ODE it must then by given by $f(t)=s+t$, as desired.