I'm new to connections and I'm going over the page (https://mathworld.wolfram.com/VectorBundleConnection.html) in which they state the following
For example, the trivial bundle $E=M\times \Bbb R^k$ admits a flat connection since any bundle section $s$ corresponds to a function $s:M\to \Bbb R^k$. Then setting $\nabla s=ds$ gives the connection. Any connection on the trivial bundle is of the form $\nabla s=ds+s \otimes \alpha$, where $\alpha$ is any one-form with values in $\text{Hom}(E,E)=E^*\otimes E$, i.e., $\alpha$ is a matrix of one-forms.
I understand from here that if $s: M \to E= M\times \Bbb R^k$ is a section of $E$, then this is identifiable with a map $s :M \to \Bbb R^k$. Now $ds$ is a bit ambiguous, but if I'm not mistaken this is under the identification $s=(s_1,\dots,s_k)$ with $s_i : M \to \Bbb R$ just $$ds=(ds_1,\dots,ds_k).$$
The second part of the paragraph is what confuses me a bit, they state that any connection on a trivial bundle is of the form $$\nabla s=ds +s\otimes \alpha$$ for $\alpha \in \Gamma(T^*M\otimes \text{Hom}(E,E))$. Where is this coming from, why is any connection given by just giving a matrix of $1$-forms?
For example on the trivial bundle $M \times \Bbb R^2$ over $M$, will something like $$\alpha = \begin{bmatrix}dx&0\\ 0&dy\end{bmatrix}$$
immediately give me a connection on $M \times \Bbb R^2$?
A follow up to this would be that on a smooth manifold with a rank $k$ vector bundle $E \to M$, given a trivializing open cover $\{U_i\}$, can I define connections $\nabla_i$ on $E|_{U_i} \cong U_i \times \Bbb R^k$ and patch these together to get a connection on $E$ even if it isn't trivial itself?
First question: the space of connections on a vector bundle is an affine space, whose translation space consists of tensors of the appropriate type. What this means, more precisely, is that:
Now, assume that $E = M\times \Bbb R^k$ is trivialized. Your understanding of what ${\rm d}s$ means here is correct: we identify $s$ with a mapping $M\to \Bbb R^k$, and so ${\rm d}s_p\colon T_pM \to \Bbb R^k$ eats $v\in T_pM$ and spits out ${\rm d}s_p(v) \in \Bbb R^k$.
The same thing happens for $\nabla s$: it takes $v\in T_pM$ and outputs $\nabla_vs \in \Bbb R^k$. More explicitly, if $X$ is a vector field on $M$, then $\nabla_Xs$ is a section of $E$, which we identify with a function $M\to \Bbb R^k$. Evaluate at a point $p$, and you essentially obtain an element in $\Bbb R^k$. With this identification in place, the standard flat connection ${\rm D}$ in $E$ is just ${\rm d}$. In its full ugliness, $({\rm D}_Xs)(p) = (p, {\rm d}s_p(X_p))$, with ${\rm d}s$ as explained in the previous paragraph.
The difference of the two connections $\nabla$ and ${\rm d}$ is a $C^\infty(M)$-bilinear mapping $$\mathfrak{X}(M)\times C^\infty(M,\Bbb R^k) \to C^\infty(M,\Bbb R^k),$$which eats a vector field $X$ and results in a $C^\infty(M)$-linear mapping $C^\infty(M,\Bbb R^k) \to C^\infty(M, \Bbb R^k)$. This is nothing else but a $1$-form valued in $\mathfrak{gl}_k(\Bbb R^k)$.
Second question: does $$\alpha = \begin{bmatrix} {\rm d}x & 0 \\ 0 & {\rm d}y\end{bmatrix}$$define a connection in $E = M\times \Bbb R^2$? No, because the entries of this matrix should be $1$-forms defined on $M$, not on $\Bbb R^2$. Let's say you were considering $$\alpha = \begin{bmatrix} \alpha_1 & 0 \\ 0 & \alpha_2 \end{bmatrix}$$for $\alpha_1$ and $\alpha_2$ $1$-forms on $M$ instead. In this case, identifying a section $s$ of $E$ with a function $M\to \Bbb R^2$ and writing it as $s=[f\,g]^{\rm T}$, we would set
$$\begin{split} \nabla_Xs &= {\rm d}s(X) + \begin{bmatrix} \alpha_1(X) & 0 \\ 0 & \alpha_2(X) \end{bmatrix} \begin{bmatrix} f \\ g \end{bmatrix} \\ &= \begin{bmatrix} {\rm d}f(X) \\ {\rm d}g(X) \end{bmatrix} + \begin{bmatrix} \alpha_1(X)f \\ \alpha_2(X)g \end{bmatrix} \\ &= \begin{bmatrix} {\rm d}f(X) + \alpha_1(X)f \\ {\rm d}g(X) + \alpha_2(X)g\end{bmatrix}.\end{split}$$ Exercise: compute the curvature of this connection "by hand".
Third question: yes, you can patch-up local connections to define a global one, provided they are compatible on overlaps. Just like for tensor fields. But the difference here is that the transformation law for connections has a nonlinear term. For nontrivial bundles, we choose a local frame $\mathfrak{e} = (e_1,\ldots, e_k)$ and write $\nabla e_j = \sum_{i=1}^k \omega_{ij}\otimes e_i$. If $\omega$ is the connection $1$-form relative to a frame $\mathfrak{e}$ and $\bar{\omega}$ is the connection $1$-form relative to a second frame $\bar{\mathfrak{e}}$, then $\bar{\omega} = A^{-1}\omega A + A^{-1}{\rm d}A$, where $A$ is the transition matrix given by $\bar{\mathfrak{e}} = \mathfrak{e}A$.