In my lecture notes, I read that a graph G which has vertices whose average degree is at most $d$ is not $d + 1$ colorable. This seems counter-intuitive to me.
I have tried examples with various graphs and cycles between them (so each vertex has degree 2 and hence average degree is 2) to find a case to support the claim. But I can't seem to find any.
Is this a mistake in the lecture notes or am I just not seeing something obvious?
There is no relationship between chromatic number and average degree.
On one hand, a graph with chromatic number $2$ can have arbitrarily large average degree. Just consider the complete bipartite graph $K_{n,n}$: this has average degree $n$, but chromatic number $2$.
On the other hand, you can take any graph with chromatic number $k$ and reduce its average degree to an arbitrarily small number by adding many isolated vertices. If your starting graph has $n$ vertices, it has at most $\binom n2$ edges. By adding $t\binom n2 -n$ more isolated vertices, you reduce the average degree to at most $\frac{2\binom n2}{t \binom n2} = \frac2t$, which can be made as small as you want.