Covariant derivative for higher rank tensors

Recently i have started to study tensors, but i have a problem with covariant differentiation. In every textbook i have read so far, there is a formula like $A^{ij}_{;k}=\frac{\partial{A^{ij}}}{\partial{x^k}}+A^{hj}\Gamma^i_{hk}+A^{ih}\Gamma^j_{hk}$ to calculate covariant derivative for a tensor with two contravariant ranks, but none of them has explained the way which results in such formula. Now suppose we have a two dimensional space with basis vectors $\vec{e}_1$ and $\vec{e}_2$. As you know we can build a tensor with two contravarint ranks which can be written in component form as $T = T^{11}(\vec{e}_1 \bigotimes \vec{e}_1)+T^{12}(\vec{e}_1 \bigotimes \vec{e}_2)+T^{21}(\vec{e}_2 \bigotimes \vec{e}_1)+T^{22}(\vec{e}_2 \bigotimes \vec{e}_2)$. Is there any way to reach from component form of a tensor to $A^{ij}_{;k}=\frac{\partial{A^{ij}}}{\partial{x^k}}+A^{hj}\Gamma^i_{hk}+A^{ih}\Gamma^j_{hk}$ ? In other words, is this possible to differentiate like $\frac{\partial{T^{ij}(\vec{e}_i \bigotimes \vec{e}_j)}}{\partial{x^k}}=\frac{\partial{T^{ij}}}{\partial{x^k}}(\vec{e}_1 \bigotimes \vec{e}_1)+T^{ij}(\frac{\partial{\vec{e}_i}}{\partial{x^k}} \bigotimes \vec{e}_j)+T^{ij}(\vec{e}_i \bigotimes \frac{\partial{\vec{e}_j}}{\partial{x^k}})$ ?

I'm not sure I understand your question thoroughly, but I'm assuming you're looking for a proof of the formula with the $\Gamma$ symbols that you proposed. Yes, it is actually easy to prove it starting from the definition of connection or covariant derivative of a tensor field. We define $\nabla_X$, the "covariant derivative (of a tensor field) with respect to the vector field $X$" as a map that takes in a $(p,q)$-tensor field and spits out a $(p,q)$-tensor field, and follows these rules:

1. For all scalar fields $f$ (i.e. when $p=q=0$), $\nabla_X f = Xf$;
2. For all $(p,q)$-tensor fields $T,S$, $\nabla_X (T + S) = \nabla_X T + \nabla_X S$;
3. If $T$ is a $(p,q)$-tensor field, and $Y_1, \dots, Y_q$ are vector fields, and $\omega^1,\dots,\omega^p$ are covector fields, then $$\begin{split} \nabla_X(T(\omega^1, \dots, \omega^p, Y_1, \dots, Y_q)) =&\ (\nabla_X T)(\omega^1, \dots, \omega^p, Y_1, \dots, Y_q)+\ \\ &+ T(\nabla_X\omega^1, \dots, \omega^p, Y_1, \dots, Y_q) + \cdots + \\ &+ T(\omega^1, \dots, \nabla_X\omega^p, Y_1, \dots, Y_q) +\ \\ &+ T(\omega^1, \dots, \omega^p,\nabla_X Y_1, \dots, Y_q) + \dots + \\ &+ T(\omega^1, \dots, \omega^p, Y_1, \dots, \nabla_XY_q) \end{split}$$ which is similar to the formula you propose at the end of your question, and acts as a "Leibniz rule" for $\nabla_X$;
4. If $f,g$ are scalar fields and $X,Y$ are vector fields, then $\nabla_{fX+gY} T = f \nabla_X T + g \nabla_Y T$, for all tensor fields $T$.

Now, if $X,Y$ are vector fields, we may write $$X = X^i \frac{\partial}{\partial x^i}, \qquad Y = Y^m \frac{\partial}{\partial x^m}$$ having chosen a chart $(U,x)$ in which to work. Then the covariant derivative may be expanded as $$\nabla_X (Y) = \nabla_{X^i \frac{\partial}{\partial x^i}} \left(Y^m \frac{\partial}{\partial x^m} \right) $$ and using rules 3. and 4. above $$\nabla_X (Y) = X^i \nabla_{ \frac{\partial}{\partial x^i}} \left(Y^m \right) \frac{\partial}{\partial x^m} + X^i Y^m \nabla_{ \frac{\partial}{\partial x^i}} \left(\frac{\partial}{\partial x^m}\right) $$ But $Y^m$ is a scalar field, so the action of $\nabla_{ \frac{\partial}{\partial x^i}}$ corresponds to the action of $\frac{\partial}{\partial x^i}$; moreover, the object $\nabla_{ \frac{\partial}{\partial x^i}} \left(\frac{\partial}{\partial x^m}\right)$ is a vector field, and as such may also be expanded in terms of our basis as $$\nabla_{ \frac{\partial}{\partial x^i}} \left(\frac{\partial}{\partial x^m}\right) = \Gamma^q_{mi} \frac{\partial}{\partial x^q} $$ for some scalar fields $\Gamma^q_{mi}$. In total, then, $$\nabla_X (Y) = X^i \frac{\partial Y^m}{\partial x^i} \frac{\partial}{\partial x^m} + X^i Y^m \Gamma^q_{mi} \frac{\partial}{\partial x^q} $$ If we pick $X = \frac{\partial}{\partial x^k}$, the formula reduces to $$\nabla_{\frac{\partial}{\partial x^k}} (Y) = \frac{\partial Y^m}{\partial x^k} \frac{\partial}{\partial x^m} +Y^m \Gamma^q_{mk} \frac{\partial}{\partial x^q} $$ which, when taking components (and cleaning up indices), is just your formula in the simpler case of a 1-contravariant tensor field $Y$: $$Y^a_{;k} \frac{\partial}{\partial x^a} := (\nabla_{\frac{\partial}{\partial x^k}} (Y))^a \frac{\partial}{\partial x^a} = \left(\frac{\partial Y^a}{\partial x^k} + Y^m \Gamma^a_{mk}\right) \frac{\partial}{\partial x^a}$$

I leave it to you to generalize this to the case of 2-contravariant tensor fields $A$. The general formula for $(p,q)$-tensor fields is pretty daunting: $$\begin{split} \nabla_X \left( T \right) &= \nabla_X \left( T^{i_1 \cdots i_p}_{j_1\cdots j_q} \frac{\partial}{\partial x^{i_1}} \otimes \cdots \otimes \frac{\partial}{\partial x^{i_p}} \otimes \mathrm{d}x^{j_1} \otimes \cdots \otimes \mathrm{d}x^{j_q} \right) \\ &= X^k \left[ \frac{\partial T^{i_1 \cdots i_p}_{j_1\cdots j_q}}{ \partial x^k} + \sum_{n=1}^p T^{i_1 \cdots \alpha_n \cdots i_p}_{j_1\cdots j_q} \Gamma^{i_n}_{\alpha_nk} - \sum_{m=1}^q T^{i_1 \cdots i_p}_{j_1\cdots \beta_m \cdots j_q} \Gamma^{\beta_m}_{i_m k} \right] \bigotimes_{\ell=1}^p \frac{\partial}{\partial x^{i_\ell}} \otimes \bigotimes_{h=1}^q \mathrm{d}x^{j_h} \end{split}$$ where the indices $\alpha_n$ and $\beta_m$ are intended to be placed, respectively, in the $n$-th and $m$-th spot of the index string (where $i_n$ and $j_m$ should have been, hadn't they been displaced). In components, if $X = \partial/\partial x^k$, $$T^{i_1 \cdots i_p}_{j_1\cdots j_q;k } = \frac{\partial T^{i_1 \cdots i_p}_{j_1\cdots j_q}}{ \partial x^k} + \sum_{n=1}^p T^{i_1 \cdots \alpha_n \cdots i_p}_{j_1\cdots j_q} \Gamma^{i_n}_{\alpha_nk} - \sum_{m=1}^q T^{i_1 \cdots i_p}_{j_1\cdots \beta_m \cdots j_q} \Gamma^{\beta_m}_{i_m k} $$ So, for example, if $p = 3$ and $q = 2$, $$T^{abc}_{rs;k} = \frac{\partial T^{abc}_{rs}}{ \partial x^k} + T^{\alpha bc}_{rs} \Gamma^a_{\alpha k} + T^{a \beta c}_{rs} \Gamma^b_{\beta k} + T^{ab\gamma}_{rs} \Gamma^c_{\gamma k} - T^{abc}_{\rho s} \Gamma^\rho_{rk} - T^{abc}_{r\sigma} \Gamma^\sigma_{sk} $$