Confusion with two-mode bra-ket notation Let us consider some abstract two-mode bosonic model with a conserved total number of quanta (i.e. eigenvalue $N$ of the operator $\hat{N} = a^\dagger a + b^\dagger b$ remains constant with corresponding subspace). Thus, the entire Hilbert space splits into a sum of invariant Hilbert sub-spaces for all $N = 0, 1, 2, \dots$. Thereunder, all of the states corresponding to the subspace with $N$ quanta have a form of a superposition of the blocks like $|n, N-n \rangle$, where $n = 0, 1, \dots, N$.
Thus, as far as I figured out, the coherent state reads
$$
|\alpha, \beta\rangle = e^{-(|\alpha|^2+|\beta|^2)/2} \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\alpha^n \beta^k}{\sqrt{n! k!} }|n, k\rangle \\  \rightarrow e^{-(|\alpha|^2+|\beta|^2)/2}\sum_{m = 0}^N \frac{\alpha^m \beta^{N-m}}{\sqrt{m! (N-m)!}} |m, N-m\rangle,
$$
where I assumed $|\alpha|^2 + |\beta|^2 = N$.
Let's consider the following state $|\alpha, 0 \rangle$, where $|\alpha|^2 = N$. I'm a bit confused, how will this state look like in the subspace with $N$ quanta?
 A: You took an unfortunate left turn in your otherwise plausible reexpression of the first quadrant semi-infinite square lattice in terms of finite N+1- length diagonals thereof, stacked along the n=k line for all Ns. As the comments warn you, one hardly ever constrains conjugate (Mellin) variables among themselves, anymore than one constrains x with p in Fourier transforms in QM.
Your coherent state is
$$
|\alpha, \beta\rangle = e^{-(|\alpha|^2+|\beta|^2)/2} \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{\alpha^n \beta^k}{\sqrt{n! k!} }|n, k\rangle \\ =\sum_{N=0}^\infty  
e^{-(|\alpha|^2+|\beta|^2)/2}\sum_{n = 0}^N \frac{\alpha^n \beta^{N-m}}{\sqrt{n! (N-n)!}} |n, N-n\rangle  \\ =e^{-|\alpha|^2  (1+|z|^2 )/2}  \sum_{N=0}^\infty \alpha^N
  \sum_{n = 0}^N \frac{   z^{N-n}}{\sqrt{n! (N-n)!}} |n, N-n\rangle ,
$$
where I  defined $\beta \equiv z \alpha$, so that $|\alpha|^2 + |\beta|^2 =|\alpha|^2  (1+|z|^2 ) $.

Possibly useful picture to summarize your rearrangement: The lower right corner (SW) is the vacuum; the abscissa is n, going East, extending to infinity; and the ordinate k, increasing North, also extending to infinity.  Your inner sums are the finite diagonals SE to NW, of lengths 0, 1, 2, ... with 1, 2, 3 points each. The outer sum adds these diagonals from SW to NE, without limit.


Your question of how does the one-mode coherent state $|\alpha, 0 \rangle$ sits in $|\alpha, z \alpha \rangle$ is  answerable by inspection of the lattice picture; as $z\to 0$, the only surviving term with vanishing exponent of the second (finite) sum is the topmost sum point at the beginning of the diagonal (n=N), $ \frac{1}{\sqrt{N! }} |N, 0 \rangle $,  restoring the standard one-mode coherent state: plug it in.
