How to go from this to this formula $$ \ddot{x}^{i}=\ddot{A}^{i}_{j}y^{i}+ 2\dot{A}^{i}_{j}\dot{y}^{i}+A^{i}_{j}\ddot{y}^{i}+\ddot{C}^{i}$$ knowing that $$ \dot{A}A^{-1}x=\omega \times x$$ $$e_{i}'=A^{i}_{j}e_{i}$$ $$r=x^{i}e_{i} \,\,\,\,\,\,\,\,\,\, v'=y^{i}e'_{i}$$ $$\dot{x}^{i}=\dot{A}^{i}_{j}y^{i} + A^{i}_{j}\dot{y}^{i}+ \dot{C}^{i}$$
$$v=\dot{x}^{i}e_{i}\,\,\,\,\,\,\,\,\,\, v'=\dot{y}^{i}e'_{i}$$
to
$$a=a'+\dot{\omega} \times r' +2\omega \times v' +\omega \times (\omega \times r') + a_{0}$$
The first observation is that this problem is about writing the acceleration in a non-inertial reference frame when you know what is happening in an inertial reference frame. The second observation is that you mix components and vector formulas. The third observation is that you have many typos/mistakes in your formulas.
In the inertial reference frame the position is described by the vector $r$ with components $x^i$ along $e_i$ axes. Similarly, in the non-inertial frame, the position $r'$ is described by $y^i$ along $e'_i$. To describe the same object in the two reference frames, $r=r'+r_0$, where $r_0$ is the position of the origin of the non-inertial reference frame with respect to the origin of the inertial reference frame, and has components $C^i$. The non inertial reference frame axes also rotate with respect to the inertial reference frame. The orientation of these axes is given by a rotation matrix, with components $A^i_j$, such that $e^i=A^i_je'^j$ (note the indices in your formula are wrong). If we write the equation for position by components you get $$x^ie_i=y^ie'_i+C^ie_i=y^iA^i_je_j+C^ie_i$$ If I change the notation for the matrix term, we can write the $i$-th component as $$x^i=A^j_iy^j+C^i$$ Taking now the derivative with respect to time we get $$\dot x^i=\dot A^j_iy^j+A^j_i\dot y^j +\dot C^i$$ If we want to write in terms of vectors, $v=\dot x^ie_i$, $v_0=\dot r_0=\dot C^ie_i$, $v'=\dot y^j e'_j=\dot y^j A^j_ie_i$. The lat term we have to deal with is $\dot A^j_iy^je_i$. But we know that $e^i=A^i_je'^j$ and $r'=y^je'_j$, so we can insert $A^{-1}A=I$ to get $\dot A A^{-1}r'=\omega\times r'$ So $$v=v'+\omega\times r'+v_0$$
By taking the derivative of $\dot x^i$ formula from above with respect to time we get $$\ddot x^i=\ddot A^j_iy^j+2\dot A^j_i\dot y^j+A^j_i\ddot y^j +\ddot C^i$$ The meaning of two terms is trivial $\ddot x^ie_i=a$, $\ddot C^ie_i=a_0$. By applying the same procedure as before, two more terms can be written as $A^j_i\ddot y^je_i=a'$ and $2\dot A^j_i\dot y^je_i=2\omega\times v'$. We therefore get $$a=a'+a_0+2\omega\times v'+\ddot A^j_iy^je_i$$ To prove the last formula, we just need to get the meaning of $\ddot A^j_iy^je_i$. Let's go back to $$\dot A A^{-1}x=\omega\times x$$ Taking the time derivative you get $$\ddot A A^{-1}x+\dot A\dot{A^{-1}}x+\dot A A^{-1}\dot x=\dot\omega\times x+\omega\times\dot x$$ Rearranging the terms, you get $$\ddot A A^{-1}x=\dot\omega\times x-\dot\omega\times x+\dot\omega\times x-\dot A\dot{A^{-1}}x$$ The derivative of $A^{-1}$ is $\dot{A^{-1}}=-\dot A A^{-2}$, so $$\ddot A A^{-1}x=\dot\omega\times x+\dot A\dot AA^{-1}A^{-1}x=\dot\omega\times x+\omega\times(\omega\times x)$$Therefore $$a=a'+a_0+2\omega\times v'+\dot\omega\times x+\omega\times(\omega\times x)$$