I find the following paragraph in the book of Manifolds and differential geometry by Jeffrey M.Lee (section 12.2 Connection forms page 506):
Let $\pi : E \longrightarrow M $ be a rank r vector bundle with a connection $\nabla$. Recall that a choice of a local frame field over an open set $U \subset M $ is equivalent to a trivialisation of the restriction $\pi_U : {E|}_U \longrightarrow U . $ Namely, if $ \phi = (\pi, \Phi)$ is such a trivialisation over U, then defining $e_i(x)= \phi ^{-1}(x,e_i)$, where $(e_i)$ is the standard basis of $\mathbb{F}^n$, we have a frame field $(e_1,...,e_k).$ We now examine the expression for a given covariant derivative from the viewpoint of such local field. It is not hard to see that for every such frame field there must be a matrix of 1-forms $\omega = {(\omega_j^i)}_{1≤i,j≤r}$ such that for $X \in \Gamma(U,E)$ we may write $$\nabla_Xe_j =\sum\limits_{i=1}^k {\omega_j^i}(X)e_i$$
My question is how can prove the existence of such a matrix of 1-forms ?
We know that $\nabla_Xe_i$ is a smooth section of $E$, and working locally over a frame $(U,e_1,\ldots, e_k)$ as you are we have $$ \nabla_Xe_i=\sum_{j=1}^n \omega^j_i(X)e_j$$ where $\omega^j_i:\mathfrak{X}(U)\to C^\infty(U).$ It is easy to see using $\nabla_{aX+bY}=a\nabla_X+b\nabla_Y$ that $\omega^j_i$ is linear, but more is true. $\omega^j_i$ is tensorial because $\nabla_{fX}=f\nabla_X$ so that $\omega^j_i(fX)=f\omega^j_i(X)$. Hence, $\omega^j_i$ is a smooth $(0,1)-$tensor field over $U$. That is, the $(\omega_j^i)$ form a $k\times k$ array of smooth $1-$forms over $U$.
Edit: This is to address the question brought up in the comments.
You want to express the matrix $(\omega_j^i)_{i,j}$ defined locally as a section of a bundle of endomorphisms. It is possible to define vector bundle valued differential forms. Actually, this is not so hard. In a local trivialization of $E$ given by $(U,e_1,\ldots, e_r)$ an $E-$valued differential $k-$form is an expression of the form $\eta=\eta^1\otimes e_1+\cdots +\eta^r\otimes e_r$ for each $\eta^i$ a $k-$form. That is, we have a $k-$form in front of each frame element, so that given $X_1,\ldots, X_k$ vector fields on the same open neighborhood, $$ \eta(X_1,\ldots, X_k)=\sum_{i=1}^r \eta^i(X_1,\ldots,X_k)e_i$$ defines a section of $E$ over $U$. With all this being said, if $E$ is locally trivial with $e_1,\ldots, e_r$ a local frame, then $\operatorname{End}(E)$ is also trivialized, with frame given by the standard matrices $E_{i,j}$ (all zeroes except a $1$ in the $(i,j)$ position). That is, $E_{i,j}$ represents the transformation sending $e_j\mapsto e_i$ and all other $e_k\mapsto0$. Then a $\operatorname{End}(E)-$valued $k-$form is exactly a sum of the form $$ \sum_{i,j}\underbrace{\eta^{i,j}}_{\in \Omega^k(U)}E_{i,j}= \begin{bmatrix} \eta^{1,1}&\cdots& \eta^{1,r}\\ \vdots&\ddots&\vdots\\ \eta^{r,1}&\cdots& \eta^{r,r} \end{bmatrix}.$$ In the case where $k=1$, this is called an $\operatorname{End}(E)-$valued $1-$form. Note that this is a special case of the general construction for vector bundles, since $\operatorname{End}(E)$ is a vector bundle. This is what the author is referring to as $\mathcal{A}^1(-,\operatorname{End}(E))$.