Let me give few ideas that might help, although I do not have a complete proof (yet).
Suppose that $G$ has intersecting odd cycles and that $G$ has no $K_5$ as subgraph. We shall prove that $\chi(G)\leq 4$.
Let us suppose that $G$ contains no triangle. Let $C$ be the smallest odd cycle of $G$ (which has at least five vertices). $G-C$ is bipartite with parts $A$ and $B$. Note that every vertex of $A\cup B$ has at most two neighbors in $C$, because $G$ has no triangles and $C$ is a minimum odd cycle of $G$ (having at least three neighbors in $C$, imply on two neighbors at distance at least four in $C$, since $G$ has no triangles, and thus on a smaller odd cycle). Consequently, one can extend a coloring of $C$ with colors 1,2 and 3, to a coloring of $G$ by coloring the whole set $A$ with color 4 and coloring each vertex of $B$ with a color in the set ${1,2,3}$ that does not appear in its neighborhood in $C$.
The problem is then how to solve the case when $G$ has a triangle $C$? I do not know yet how to extend a 3-coloring of the triangle to a 4-coloring of $G$. Of course, the hypothesis of having intersecting odd cycles and having no $K_5$ as subgraph are important now.
For instance, one can prove that if $G-C$ has parts $A$ and $B$ and $A_i$ (resp. $B_i$) corresponds to the set of vertices in $A$ (resp. $B$) with exactly $i$ neighbors in $C$, for every $i\in\{0,\ldots,3\}$, then $A_3\neq \emptyset$ and $B_3\neq\emptyset$, as otherwise it is possible to find a 4-coloring of $G$ by a similar argument as in the previous case. Supposing that $A_3$ and $B_3$ are both non-empty lead us to deduce that $A_3\cup A_2\cup B_3 \cup B_2$ is an independent set, as otherwise one may find a $K_5$ or two disjoint triangles. Since we do not have two disjoint triangles, $A_3\cup B_1$ and $B_3\cup A_1$ are also independent sets. Moreover, there is no edge $a_1b_1$ such that $a_1\in A_1$, $b_1\in B_1$ and $a_1$ and $b_1$ have the same neighbor in $C$.
Yet, I cannot color $G$. The vertices with no neighbor in $C$ are an issue.
Best Answer
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.