I am struggling to understand how this derivative works (differentiating a functional from calculus of variations):
$$\frac{\partial}{\partial\alpha}f(y(x)+\alpha\eta(x), y'(x)+\alpha\eta'(x), x)=\eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'}$$
I am not sure why we need to add two terms in the final result.
Let's assume, that we have function $t=f(x,y,z)=f(x(u,v, \alpha), y(u,v, \alpha),z(u,v, \alpha))$, then chain rule with respect, for example, $\alpha$ will be
$$\frac{\partial f}{\partial \alpha}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial \alpha}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial \alpha}$$ In your case it's enough to see that $\frac{\partial x}{\partial \alpha} = \eta$, $\frac{\partial y}{\partial \alpha} = \eta'$ and $\frac{\partial z}{\partial \alpha} = 0$.
To make things looking exactly as in question, let's consider $t=f(y,y',y'')=f(y(u,v, \alpha), y'(u,v, \alpha),y''(u,v, \alpha))$ Then $$\frac{\partial f}{\partial \alpha}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha}+\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial \alpha}+\frac{\partial f}{\partial y''}\frac{\partial y''}{\partial \alpha}$$ and $\frac{\partial y}{\partial \alpha} = \eta$, $\frac{\partial y'}{\partial \alpha} = \eta'$ and $\frac{\partial y''}{\partial \alpha} = 0$.