Omitting technical details, if we want to find a local extremum $x_1$ of a functional $f:E_1\to\mathbb R$ on a Banach space $E_1$ with constraint $g(x_1)=0$, where $g:E_1\to E_2$ and $E_2$ is another Banach space, we may solve $${\rm D}f(x_1)+\lambda{\rm D}g(x_1)=0\tag1,$$ where $\lambda\in E_2'$.
Can we incorporate $g(x_1)=0$ and $(1)$ into one equation $${\rm D}\mathcal L(x_1,\lambda)=0\tag2,$$ where $$\mathcal L(x_1,\lambda):=f(x_1)+\lambda g(x_1)$$ as it is explained on Wikipedia in the Euclidean space case?
It should be true, but the Fréchet derivative in $(2)$ involves Fréchet derivatives of functions on $E_2'$ which seems to be quite complicated and maybe some identification is necessary.
Lagrangian function in any space (finite on infinite dimension ) is
$$\mathcal L(x_1,\lambda):=f(x_1)+ \langle \lambda , g(x_1) \rangle $$
Yes. You need to consider $weak-star$ topology on $E'_2$. Then in this case you have $E''_2 = E_2$
In general when you assume $E'_2$ equipped with operator-norm The derivative of lagrangian with respect to $\lambda$ at a fix point $\bar \lambda$, is a bounded linear functional, say $\phi : E'_2 \to R$ defined by $\phi(z) = \langle z ,g(x) \rangle $. Setting this $\phi$ equals zero gives you again $g(x)=0$, so no problem in this case too. Assuming $weak-star$ topology on $E'_2$ gives you $\phi : E_2 \to R$, which is standard definition.
Then derivative with respect to $x$ gives equation (1). While the derivative with respect to $\lambda$ gives you the constraint $g(x)=0$.