An all-to-brief review
Let's cover the basics: you take a set of points $\mathcal M$ and outfit it with a set of scalar fields $\mathcal S \subseteq (\mathcal M \to \mathbb R)$ which obey a closure axiom under the smooth functions $C^\infty(\mathbb R^n, \mathbb R),$ reinterpreted pointwise as functions $\mathcal S^n \to \mathcal S$ via a lifting operator $\mathcal L$ where $$\mathcal L_n f(s_1,\dots s_n) = p\mapsto f\big(s_1(p),\dots s_n(p)\big).$$This gives us constant fields $\mathcal L_0 1$ and pointwise $(+)$ and $(*)$ as $\mathcal L_2 (+),\mathcal L_2 (*),$ on those fields: and you say that you've seen that this induces a topology where a subset of $\mathcal M$ is called "closed" if it is the kernel of a scalar field, or "open" if its complement is closed; on this topology all of the fields $\mathcal S$ are in fact continuous maps. Then there's the "coordinate map" axiom which says that the space is $D$-dimensional: for every point $p\in\mathcal M$ there is an open set containing $p$ where all scalar fields can be expressed in terms of a set of coordinate fields $c_1, c_2, \dots c_D$ as smooth functions, $$s = \mathcal L_D \sigma(c_1, \dots c_D).$$
Now you have also come to the fundamental smart thing about vector fields, that $\mathcal V$ is the space of derivations, linear maps $\mathcal S \to \mathcal S$ obeying the generalized Leibniz law that if $V \in \mathcal V$ then $$V\big(\mathcal L_n f(s_1,\dots s_n)\big) = \sum_{k=0}^n \mathcal L_n f_{(i)}(s_1, \dots s_n) * V s_i,$$ where $f_{(i)}$ is the partial derivative of $f \in C^\infty(\mathbb R^n, \mathbb R)$ with respect to its $i$th argument. And then the covector fields $\mathcal V^\dagger$ are linear maps $\mathcal V \to \mathcal S.$ While at first this definition of "vector fields" seems like a strange definition, it retroactively makes sense because looking at that equation and the one which immediately precedes it, we see that locally every vector field has a set of $D$ scalar-field components $v_i = V c_i$ just like you'd expect, leaving $V = \sum_i v_i \partial_i.$
And now we can define valence-$[m, n]$ tensor fields as multilinear maps from $\big(\mathcal V^\dagger\big)^m\times\big(\mathcal V\big)^n \to \mathcal S,$ and with a bunch more axioms we get an abstract-index notation complete with outer products and contractions. This also gives us directly the Lie bracket $[U, V] = U \circ V - V \circ U,$ and the gradient covector for any scalar field, $\nabla_\bullet s = V \mapsto V s.$ We don't need a connection to have these more-basic aspects.
OK, now why are vectors different?
You can interpret $\sum_i v_i \partial_i$ as the term from the transport equation, $\partial_t \rho + \vec v \cdot \nabla \rho = -\nabla \cdot \vec J + \Omega,$ describing essentially "an infinitesimal box flows downstream, the rate of change of some conserved stuff in the box is equal to the divergence of flows through the nearby boxes, plus any creation or annihilation of stuff (or flow into the system of interest from some other place). In particular this idea of "a box flows downstream" means that we're trying to analyze $\big[\rho(\vec r + d\vec r, t + dt) - \rho(\vec r, t)\big] / dt$ which of course is just the left-hand side with $d\vec r/dt = \vec v.$
Well, scalars and vectors-in-flat-space can be transported without their identity becoming ambiguous, but vectors-in-curved-space can't: imagine that you are pointing East as you walk to the North Pole, you are now pointing South (as you must, at the North Pole), and the actual longitude that you're pointing depends on how you walked there (with the straight shot, you're pointing at something 90 degrees to the East; but if you first walk sideways West you can point to increasingly Westward longitudes.
In fact we can directly anticipate that your solution will only be unique up to a $[1,2]$-valence tensor, because in practice you're going to form $w^\alpha = u^\beta \nabla_\beta v^\alpha$ and so you're taking these two vector fields $u^\beta$ and $v^\alpha$ and forming a new one $w^\alpha$, and mapping two vectors to a vector is the same as mapping two vectors and a covector to a scalar, which is a $[1,2]$-tensor field.
We can derive it quite directly; note that if you have two connections $\nabla_\alpha$ and $\hat \nabla_\alpha$ that agree on how to treat scalar fields (our universal gradient above) then the operator $\delta_\alpha = \nabla_\alpha - \hat \nabla_\alpha$ must satisfy $\delta_\alpha s = 0$ for all scalar fields $s$, hence $\delta_\alpha (s v^\beta) = s \delta_\alpha v^\beta.$ And that's a multilinear map from $\mathcal T^\beta \to \mathcal T_\alpha^\beta$ which makes it precisely a [1, 2]-tensor. In some sense this freedom means that you can use the existing connection but allow a tensor to make slight tweaks to any transporting vectors, arbitrarily. And then we've got these ideas like "torsion" which allow us hopefully to push past that.