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.
So, since you asked in the comments above, here is your confirmation.
Maybe I can add, that since you have sections $s_i$ on $U$ in each of your principal bundles, they are all trivializable over that same $U$, consequently each of the associated vector bundles is trivializable over this $U$.
On the other hand, $C^\infty(U,V_1\otimes\cdots\otimes V_n)$ is a section space over $U$ in the trivial bundle $U\times(V_1\otimes\cdots\otimes V_n)$ and applying the inverse of the respective trivialization to each factor should simply give you the isomorphism
$$
C^\infty(U,V_1\otimes\cdots\otimes V_n)\to\Gamma(U,E_1\otimes\cdots\otimes E_n).
$$
The (inverse) trivilization in each factor is then, as you remarked in the footnote, given by $v\mapsto[s_i(m),v]$.
Best Answer
Are you sure this is true?
Consider the following: take $M = S^1$ to be the unit circle and take $B \to M$ to be trivial line bundle with fiber $\mathbb R^1$. The space of smooth global sections of $B$ is then just the space of smooth functions on $S^1$. Equip $S^1$ with the usual metric on the circle and equip $B$ with the Euclidean metric. Take the connections involved to be the trivial ones.
Now let $f : S^1 \to [-1,1]$ be the function induced by $x \mapsto \sin(2\pi x)$. This function is smooth, and thus a smooth section of $B$. Note that the absolute value function $|\cdot| : [-1,1] \to \mathbb R$ is continuous. By the Weierstrass approximation theorem there exists a sequence of polynomials $p_m$ that converges uniformely to $|\cdot|$ on $[-1,1]$. The composition $s_m = p_m \circ f$ is thus a sequence of smooth sections of $B$, that converges to $s(x) = |\sin(2\pi x)|$, which is not a smooth section of $B$ (its derivative is discontinuous at the point corresponding to the zero $\pi$ - just plot the graph of the function to see it).
I haven't verified the details, so it is possible that the sequence $(s_m)$ does not converge in the seminorm required (basically due to the discontinuity of the first derivative of the limit). If so, then replacing the absolute value by a function with continuous first, but not second, derivatives should give a counterexample.
In general I believe one obtains the space of $L^2$ sections of $B$ by taking the completion with respect to the seminorm you defined, and not the space of smooth sections.