How to prove $nablavec{V}$ is a tensor without transformation properties?

One can show that the covariant derivative $(nabla mathbf{v})^alpha_{~~beta}equiv v^alpha_{~~;beta}$ transforms like a $(1,1)$-tensor without using properties of the christoffel symbol but in order to do so one needs to start from a different expression for the covariant derivative: the following holds for the total differential of a vector field $mathbf{v}=v^mu(q)mathbf{e}_mu$: $$dmathbf{v}=v^mu_{~~;nu}dq^numathbf{e}_mu. tag{1}$$

The total derivative is a physical quantity and therefore is required to be invariant under coordinate transformations. In the following we use the two coordinate systems ${alpha^mu,beta^mu}$ with basis vectors ${mathbf{a}_mu,mathbf{b}_mu}$. The total differential in the different systems reads: $$dmathbf{v}=bar v^mu_{~~;nu}dalpha^numathbf{a}_mu=v^mu_{~~;nu}dbeta^numathbf{b}_mu. tag{2}$$ Plugging standard identities for the transformation of basis vectors and coordinate differentials $$dbeta^nu=Lambda^nu_{~~mu} dalpha^mu=barLambda_mu^{~~nu} dalpha^mu,tag{3a}$$ $$mathbf{b}_mu=Lambda_mu^{~~nu}mathbf{a}_nu=barLambda^nu_{~~mu}mathbf{a}_nu,tag{3b}$$ into eq. (2) results in $$dmathbf{v}=v^mu_{~~;nu}(barLambda_kappa^{~~nu} dalpha^kappa)(barLambda^lambda_{~~mu}mathbf{a}_lambda)= barLambda^lambda_{~~mu}barLambda_kappa^{~~nu}v^mu_{~~;nu}dalpha^kappamathbf{a}_lambda=barLambda^mu_{~~lambda}barLambda_nu^{~~kappa}v^lambda_{~~;kappa}dalpha^numathbf{a}_mutag{4}$$ and therefore $$bar v^mu_{~~;nu}=barLambda^mu_{~~lambda}barLambda_nu^{~~kappa}v^lambda_{~~;kappa} quadtext{q.e.d.}tag{5}$$

Eq. (5) is the explicit transformation of a mixed $(1,1)$ tensor of rank 2. The $Lambda$-tensors perform coordinate transformations between the two systems: $$Lambda^{mu}_{~~nu}=frac{partial alpha^mu}{partial beta^nu}=mathbf{a}^mu cdot mathbf{b}_nu,$$

$$barLambda^{mu}_{~~nu}=frac{partial beta^mu}{partial alpha^nu}=mathbf{b}^mu cdot mathbf{a}_nu,$$

$$Lambda^{mu}_{~~nu}=barLambda_{nu}^{~~mu}.$$