Given a $C^{k}$-manifold what is the maximal $l$ that allows to have a good definition of a $C^{l}$-vector field on that manifold? In some textbooks is given a definition with $l=k-1$ and restriction $k\ge2$, in others with $l=k$ without restriction on $k$, but there isn't a proof of one or other choice, only definition.
Is there a proof? Why the maximal $l$ is $l=k-1$ or $k=l$?
In these cases is the vector field still a derivation?
I've always had to deal with smooth manifold and so it's something new for me to consider this details.
Here is an explanation why $l=k-1$ is the right answer. Let $f: U\to U'$ be a $C^k$-smooth diffeomorphism of open subsets of ${\mathbb R}^n$, i.e. $f, f^{-1}$ are differentiable of order $k$ with continuous partial derivatives of order $k$. Note that by the inverse function theorem in ${\mathbb R}^n$ it suffices to assume that $f$ is a $C^k$-smooth bijection with invertible derivative at all points of $U$. Let $V$ be a $C^m$-smooth vector field on $U$. Then the "push-forward" of $V$ under $f$, denoted $f_*(V)$, is given by $$ W(y)= Df(x)\cdot V(x), y=f(x). $$ For instance, even if $V$ is $C^\infty$, the matrix-valued function $Df(x)$ is only $C^{k-1}$-smooth and, hence, one expects $W(y)$ to be only $C^{k-1}$-smooth. Note that if $V$ is $C^m, m=k-1$ and $f$ is $C^k$, the vector field
$$ W(y)=Df(f^{-1}(y))\cdot V(f^{-1}(y)) $$ is at least $C^m$. (Composition and product of $C^m$-smooth maps is again $C^m$).
To check that one cannot do better than $m=k-1$, consider an example with $n=1$, $k=1$: $$ f(x)=\begin{cases} x^2+x& \text{if } x\ge 0\\ -x^2+x& \text{if } x\le 0 \end{cases} $$ This is a $C^1$-smooth diffeomorphism.
Take the constant vector field $V(x)=1$. (I intentionally do now use the math notation $V=\frac{\partial}{\partial x}$ for such vector fields since, most likely, you are not familiar with this formalism.) Then $W=f_*(V)$ is given by $f'(f^{-1}(y))$, i.e. $$ W(y)=\begin{cases} \sqrt{1+ 4y}& \text{if } y\ge 0\\ \sqrt{1- 4y}& \text{if } y\le 0 \end{cases} $$ I will leave it to you to check that $W(y)$ is continuous but not differentiable, hence, you only get $m=k-1=0$-smoothness (i.e. continuity) of $W$ in this example, just as expected.
To conclude: If $M$ is a $C^k$-smooth manifold, you can define $C^{k-1}$-smooth vector fields on $M$ (as vector fields which are $C^{k-1}$-smooth in local coordinates) but not $C^k$-smooth vector fields on $M$.