It is a well known theorem that a graph is bipartite (2 colourable) if and only if all its cycles are of even length.
So if you have a graph which has an odd cycle, it will require at least 3 colours.
For instance you can find a proof here: http://books.google.com/books?id=aR2TMYQr2CMC&pg=PA18
In any case, for a simpler proof of: To a cycle of odd length, to each vertex you cannot assign one of two colours so that adjacent vertices get different colours
Say you try and walk around the cycle. Each step you take, you flip the colour. So to get back to the original vertex, you need to take an even number of steps. That is not possible with an odd cycle. So you need at least 3 colours.
Note that if the graph you have is exactly two cycles which share exactly one vertex (like in your examples), then that graph is 3 colourable.
This is because a cycle is 3 colourable, even if you pre-assign a vertex with a given colour. You colour one cycle first, then based on the colour of the common vertex, you colour the other cycle.
In general, though, determining the chromatic number of a graph is a hard problem. In fact, it is NP-Complete.
There are two key observations to make: First, the hypothesis is that each pair of odd cycles share a vertex. This includes the pair $C,K$ for every $K$ an odd cycle in $G$. Hence when we remove $C$ from $G$, each odd cycle loses a vertex, and this is where the second key observation comes in: cutting a vertex from a cycle leaves a path, and paths are no obstruction to being bipartite.
Best Answer
This was proved by Stong in "Solution to problem 11086" in the American Mathematical Monthly, which you can find at https://www.jstor.org/stable/27641938. I am unable to 'copy-paste' the short argument here, so I give the brief idea.
The $\binom{k}{2}$ is probably not sharp, so we instead should use it as a hint for the proof that we need to consider pairs of colors. The key idea is that for any two colors in a proper coloring, the vertices colored with those two colors induce a bipartite graph, and so we may 'swap' the two colors on any component of this induced graph.
Suppose we have $k$-colored the graph apart from a vertex $v$. For every two colors, either the graph induced on these two colors and $v$ is bipartite, in which case we can recolor in order to color $v$, or this induced graph has an odd cycle containing $v$. Since $v$ is in fewer than $\binom{k}{2}$ odd cycles, the first case must occur as desired.