Let me explain here an actually useful characterisation of a "tensor" that is not as oldfashioned as the one of how a "tensor" transforms under change of coordinates. In the process I hope I can clarify why the difference between two connexions is a "tensor".
I shall assume smoothness everywhere.
A connexion, the way it is defined by the Original Poster, $D:\Gamma(E)\to\Gamma(E\otimes T^*M)$ takes smooth sections of the vector bundle $E$ to differential $1$-forms taking values at sections of $E$. One can see this by recognising that $\Gamma(E\otimes T^*M)\cong\Gamma(E)\otimes_{C^\infty(M;\mathbb{R})}\Gamma(T^*M)\cong\mathrm{Hom}_{C^\infty(M;\mathbb{R})}(\mathfrak{X}(M;\mathbb{R});\Gamma(E))$, where I am identifying the $C^\infty(M;\mathbb{R})$-module of sections of the tangent bundle $\Gamma(TM)$ with vector fields of $M$, as derivations on the algebra $C^\infty(M;\mathbb{R})$ ---I do recommend Lee's Introduction to smooth manifolds, or Wald's General Relativity if one is not used to these notions.
That being said, let $s\in\Gamma(E)$ a section of $E$, $X\in\mathfrak{X}(M;\mathbb{R})$ a vector field on $M$ and $f\in C^\infty(M;\mathbb{R})$ a smooth function. By applying the definition of a connexion (as given by the Original Poster), one can readily see that
$$Ds(fX)=fDs(X)$$
and
$$Dfs(X)=X(f)s+fDs(X) \ .$$
Remark: this last property is usually a defining property for a connexion on a vector bundle when it is defined globally, as opposed to the Original Poster's local definition.
Now, if $D$ and $D'$ are two connexions defined on $E$, their difference satisfies
$$(D-D')s(fX)=f(D-D')s(X)$$
and
$$(D-D')fs(X)=f(D-D')s(X) \ .$$ This last equality ---which does not hold for any of the connexions alone--- that characterise their difference as being a "tensor"; their difference is actually a differential $1$-form taking values on the sections of $E$.
The reason why the difference of two connexions satisfies $(D-D')s(fX)=f(D-D')s(X)$ is because each contribute with an extra term $X(f)s$, $Dfs(X)=X(f)s+fDs(X)$ is a sort of Leibniz rule.
A "tensor" is just a mapping that when its arguments are multiplied by functions, it behaves as a linear mapping regarding this $C^\infty(M;\mathbb{R})$-module structure. This is not as precise as I like, but let me show an example with the metric "tensor" (a riemanninan structure $\mathrm{g}$): with vector fields $X,Y,Z\in\mathfrak{X}(M;\mathbb{R})$ and smooth function $f\in C^\infty(M;\mathbb{R})$, the riemannian metric satisfies $\mathrm{g}(X+fY,Z)=\mathrm{g}(X,Z)+f\mathrm{g}(Y,Z)$.
This type of behaviour, with respect to multiplication by functions, guarantees that these "tensors" only depend on what happens at a point. And the way they transform under diffeomorphisms (change of coordinates) can be deduced from that property.
Homework for the Original Poster: since you know how covariant and contravariant tensors transform under change of coordinates, and I claimed that the difference between two connexions is pretty much a $1$-form taking values on sections of $E$: use a local basis for sections of $E$, and another one for the vector fields of $M$, and show how the values of $Ds(X)$ and $(D-D')s(X)$ change under a change of coordinates. Hint: exploit how the connexions behave when multiplying their arguments by functions.
Best Answer
Look like the confusion is about $x^\gamma$: $x^\gamma$ is not a tensor, it's just a coordinates. As a tensor field, each component of $A$ is a function of $x^1, \cdots, x^n$, and
$$ \frac{\partial}{\partial x^\gamma} A_{\alpha \beta...}$$
is really just the partial derivative of the function $A_{\alpha\beta ...}$ with respect to $x^\gamma$. In general, $A_{\alpha\beta ... , \gamma}$ is not a tensor: this is explained in the section "covariant derivatives" in the same wiki page.