Then it gives rise to a multilinear map $\mathscr{T} : TM \otimes \cdots \otimes TM \to E$ of vector bunldes over $M$ given by $\mathscr{T}_p(X_1(p), \cdots, X_n(p)) = \omega(X_1(p), \cdots, X_n(p)) \cdot s(p)$
Where $X_1, \cdots, X_n$ are sections of $TM$, or vector fields on $M$
(Remember that an $n$-form on $M$ eats $n$ vectors at each point on the tangent space and spits a real number)
The key point to take from this is that $\Omega^k(M, E) = \Gamma(E \otimes \Lambda^k T^*M)$ should be thought as the space of $E$-valued $k$-forms on $M$ (because every element of this vector space is an alternating multilinear map $TM^{k \otimes} \to E$, which are kind of exactly what forms are, right)
So like if $E$ is the one dimensional trivial line bundle on $M$ this theory agrees with the de Rham chain complex
So if $X$ is a vector field on $M$, $D_X$ is an object of the form $Z \otimes \omega$ where $\omega$ is a 1-from on $M$. Define $\nabla : \Gamma(TM) \otimes \Gamma(TM) \to \Gamma(TM)$ by $\nabla_X Y = \omega(Y) Z$.
If $E$ is a general vector bundle on $M$ then $D : \Gamma(E) \to \Gamma(E \otimes T^*M)$. Note that $D$ satisfies $D(f s) = f \otimes Ds + df \otimes s$ where $f$ is a function on $M$, because $f \in \Omega^0(M)$. That's what 0-forms are.
The first term is $Df$, which lives in $\Gamma(E \otimes T^*M)$, scaled by a function $f \in C^\infty(M)$ using the natural $C^\infty(M)$-module structure on the space of sections on any vector bundle over $M$
And the second term is switched, as you pointed out
I have $D(X) = Z_X \otimes \omega_X$ explicitly written down in components
I then define $\nabla : \Gamma(TM) \otimes \Gamma(TM) \to \Gamma(TM)$ by $\nabla_X Y = \omega_X(Y) \cdot Z_X$. ($\omega_X$ is a $1$-form, so I can feed it a vector field $Y$ and get a scalar function on $M$ which I am scaling the vector field $Z_X$ by)
But yeah this dude $\nabla_\bullet \bullet : \Gamma(TM) \otimes \Gamma(TM) \to \Gamma(TM)$ is linear on the bottom (X) component and Leibniz on the actual (Y) component
So $D$ took a vector field and gave us another vector field and a section of the cotangent bundle, and we used them to define a connection which satisfies all the connection crap
There is no canonical way to construct one on a given manifold. There are infinitely many choices of connections
BUT
if $M$ has a Riemannian metric $g$, there is a unique connection $\nabla$ such that it is compatible with the metric i.e. $Zg(X, Y) = g(\nabla_Z X, Y) + g(X, \nabla_Z Y)$ plus another technical condition which says $\nabla_X Y - \nabla_Y X = [X, Y]$
The first condition is like saying $(f \cdot g)' = f' \cdot g + f \cdot g'$ where cdot means dot product
But that theorem (called the fundamental theorem of Riemannian geometry) is remarkable because it says any geometry coming from the connection is intrinsic to the metric geometry
If you compute $D^2(X)$, which lives in $\Gamma(TM \otimes T^*M \wedge T^*M)$ so is a (1, -2)-tensor and you feed that two vector fields $Y$ and $Z$ you should obtain the Riemann curvature $R(X, Y)Z := \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X, Y]} Z$
Which is $(1, 0)$-valued itself
But again it's a long ass computation
This kind of justifies $D^2$ to be called curvature, as it agrees with the Riemannian notion of curvature for $E = TM$
