It should be clear from the definitions that the example given is roughly the only thing that can go wrong. In other words, it is when $C^1$ does not go into $C^{0,1}$. This difference is of course because $C^1$ only cares about the points that are, in some sense, infinitely close. Whereas $C^{0,1}$ can jump across boundaries, in some sense. 

Perhaps an example would make it clear:

Let $\Omega_1$ be the half plane $x < 0$ Let $\Omega_2$ be the region $x > y^k$ for some positive integer $k$. Let $f(x,y) = y^2$ in $\Omega_2$ and $- y^2$ in $\Omega_1$. Then $f$ is clearly in $C^1(\bar{\Omega_1})$ and $C^1(\bar{\Omega_2})$. To check that it is in $C^1(\overline{\Omega_1\cup\Omega_2})$ \emph{it suffices to check that $\lim_{(x,y)\to (0,0)}f$ and $\nabla f$ agrees. In this case, both functions vanish at the origin. 

It is also very clear that $f\in C^{0,1}(\bar{\Omega_i})$ for $i = 1,2$ individually. But by the definition of the Holder norms, to compute the Holder constant of a point $z\in \Omega_2$ relative to $\Omega_1\cup \Omega2$, you have to take the sup over, say, all points within distance 1 of $z$, not just those points in $\Omega_2$. So if you take the sequence of points $(x_j,y_j) = (2j^{-k}, j^{-1})$. Note that $(x_j,y_j) \in \Omega_2$ and $(-x_j,y_j)\in \Omega^1$. The distance between them is $4j^{-k}$, while the difference of the value of the function $f$ is 

$$ | f(x_j,y_j) - f(-x_j,y_j) | = 2 j^{-2} \not\leq 4 C j^{-k} $$

if $k > 2$. Just to re-iterate, the problem happens because "points between distinct regions in the set $\Omega_1\cup \Omega_2$ are allowed to get together too close too fast." And in fact this is the only possible way for things to break. Once you put a lower bound on the speed of approach of points in $\Omega_2$ to $\Omega_1$ along the boundary, you can use continuity of the first derivative implied by $C^1$ to control how fast the the difference between the function can grow in the two regions. For example, you can compute in the above example, that you have $C^{0,\alpha}$ for every $0 < \alpha < 2/k$  , while no inclusion is possible for $\alpha > 2/k$. 

This naturally leads to the external cone condition for domains, if you want $C^1$ to always include into $C^{0,1}$.