Number of edges in $K_4$-free graphs
I think $K_4$-minor-free graphs have at most $2n - 3$ edges for $n \geq 2$, and this is tight. Here's one way of looking at it. 

It is well-known that the $K_4$-minor-free graphs are exactly the graphs of treewidth at most two, see http://en.wikipedia.org/wiki/Forbidden_graph_characterization . Standard theory on treewidth tells us that a graph of treewidth at most 2 is 2-degenerate (see http://en.wikipedia.org/wiki/Degeneracy_%28graph_theory%29 ), which means that all induced subgraphs have a vertex of degree at most two. So to count the number of edges in a $K_4$-minor-free graph, we can do the following: we find a vertex of degree at most two, and delete it. We remove one vertex, and at most two edges. What we're left with is still $K_4$-minor-free (since minor-freeness is preserved when deleting vertices), so if the graph is not yet empty then we know it is 2-degenerate, and has another vertex of degree at most two. If we keep removing vertices of degree at most two then we delete $n$ vertices until the graph is empty, and for each removed vertex we delete at most two edges. Hence the number of edges is at most $2n$. 

Now observe that when removing the second-to-last vertex, there are only two vertices left so there can be only one edge that is removed in that step. Similarly, when we've removed all but one of the vertices then there can be no edges left, so in the last step we don't remove any edges. This shows that the number of edges is at most $2n - 3$ because we save 1 in the second-to-last step, and 2 in the last step.

To see that this is tight, start with a graph on two vertices and one edge. Now repeatedly select two adjacent vertices, and add a new vertex with edges to the selected two. The constructed graph will have treewidth two (since it is a 2-tree, see http://en.wikipedia.org/wiki/K-tree ) and will therefore be $K_4$-minor-free, and the number of edges you build in this way is $2n-3$.