Let $f$ be a function of two variables, and $g(r,\theta)=f(r\cos\theta,r\sin\theta)=f(x,y)$
Notation: $\frac{\partial^{i_1+...+i_k}f(x_1,...,x_n)}{\partial x_1^{i_1}...\partial x_k^{i_k}}\equiv f_{x_1^{i_1}...x_k^{i_k}}$ for simplicity.
$\Delta f(x,y)=\frac{\partial^2 f(x,y)}{\partial x^2}+\frac{\partial^2 f(x,y)}{\partial^2 y}$
At some point in the proof of $\frac{\partial^2 g(r,\theta)}{\partial r^2}+\frac{1}{r}\frac{\partial g(r,\theta)}{\partial r}+\frac{1}{r^2}\frac{\partial^2g(r,\theta)}{\partial^2 r}=\Delta f(x,y)$ we write $$f_{xx}(r\cos\theta,r\sin\theta)=\cos\theta \frac{\partial}{\partial r}[g_1(r,\theta)]-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}[g_1(r,\theta)]$$
after having calculated
$$g_1(r,\theta):=f_x(r\cos\theta,r\sin\theta)=\cos\theta g_r(r,\theta)-\frac{\sin\theta}{r}[g_{\theta}(r,\theta)]$$
I don't understand why that's true. And the same is written for $f_{yy}$.
Your last equation is valid for any function $f$ of the cartesian coordinates and the corresponding function $g$ of the polar coordinates. Thus, in particular it's valid for the pair $f_x,g_1$. Substituting $f_x$ for $f$ and $g_1$ for $g$ yields your penultimate equation.