Following <PERSON>'s urging (he commented on your question above) and the fact that I recently successfully applied similar ideas to study band-structure degeneracies in crystalline solids (see here https://arxiv.org/abs/1808.07469), let me perhaps make some thoughts/comments on your questions (even though I don't answer your questions per se).
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First, the homotopy groups that you ask for, are relatively easy to find. All you need is that for a coset space (or fiber bundle) $B = E/F$ there is a long exact sequence of homotopy groups $$\cdots \to \pi_p(E) \to \pi_p(B) \to \pi_{p-1}(F) \to \pi_{p-1}(E) \to \cdots .$$ In this sequence, each arrow is a group homomorphism. The "exactness" means that the image of any arrow is exactly equal to the kernel of the next arrow. The "long" simply states that the sequence is infinite (or 'semi-infinite' since zeroth homotopy groups terminate the sequence to the right).
As for your example, let's say that you want to know the first and the second homotopy group $$\pi_1[U(N_1+N_2+\ldots + N_\ell)/U(N_1)\times U(N_2)\times \ldots \times U(N_\ell)] \equiv H.$$ $$\pi_2[U(N_1+N_2+\ldots + N_\ell)/U(N_1)\times U(N_2)\times \ldots \times U(N_\ell)] \equiv G.$$ Of course, it is assumed that each $N_i \geq 1$.
To find it, first not that the second homotopy group $$\pi_2[U(N_1+N_2+\ldots + N_\ell)] = \mathbb{0}$$ is trivial (as is true for all Lie groups). 490
Furthermore, we find $$\pi_1[U(N_1)\times U(N_2) \times \ldots \times U(N_\ell) ] = \pi_1[U(N_1)]\oplus \pi_1[U(N_2)] \oplus \ldots \oplus \pi_1[U(N_\ell)]= \mathbb{Z}^\ell$$ where each $\mathbb{Z}$ counts the winding of the $U(1)$ phase of the determinant of the corresponding $U(N_i)$. We also need to use that $\pi_1[U(N_1+N_2 + \ldots + N_\ell)] = \mathbb{Z}$, too. Finally, since unitary group as a manifold is path-connected, its zeroth homotopy group is trivial.
The relevant segment of the long exact sequence, starting with $\pi_2(E)$ and ending with $\pi_0(E)$, leads to $$\ldots \stackrel{\varphi_a}{\longrightarrow} \mathbb{0} \stackrel{\varphi_b}{\longrightarrow} G \stackrel{\varphi_c}{\longrightarrow} \mathbb{Z}^{\ell} \stackrel{\varphi_d}{\longrightarrow} \mathbb{Z} \stackrel{\varphi_e}{\longrightarrow} H \stackrel{\varphi_f}{\longrightarrow} \mathbb{0} \stackrel{\varphi_g}{\longrightarrow} \ldots$$ where I gave the group homomorphsms some names. Clearly, the image $\textrm{im}\, \varphi_b = \mathbb{0}$ is trivial, such that by exactness also the kernel $\textrm{ker}\,\varphi_c = \mathbb{0}$. Furthermore, for any group homomorphism $\varphi:G_1\to G_2$ there is an isomoprhism $\textrm{im}\,\varphi \cong G_1/\textrm{ker}\,\varphi$, so we deduce $\textrm{im}\,\varphi_c = G = \textrm{ker}\,\varphi_d$, where at the last equation mark we used the exactness again. 677