My question is very simple, and this might be very trivial - but I appreciate the help.
Are the once continuously differentiable functions $C^1(\Omega)$ a subset of the Sobolev space $H^2(\Omega)$ containing functions with $L^2$-integrable weak derivatives up to and including order 2? In other words, does the inclusion $$ C^1(\Omega) \subseteq H^2(\Omega) $$ hold?
As Giuseppe pointed out in the comments, by passing to the derivative we can consider the simpler question $C^0 \overset ? \subset H^1$ instead, since for a $H^1$ function $f$ we have the equivalences $$f \in C^1 \iff \nabla f \in C^0$$ and $$f \in H^2 \iff \nabla f \in H^1.$$ Thus the central issue is comparing the strength of continuity to weak differentiability. It turns out that weak differentiability is (loosely) the stronger condition, so the inclusion you propose does not hold for any (non-trivial) domain.
In one dimension this is very clear-cut: weakly differentiable functions are all absolutely continuous, which is a strictly stronger condition than plain continuity. Thus when $\Omega \subset \mathbb R$ we can generate a counterexample by starting with a function which is continuous but not absolutely continuous, such as the Cantor function $c$. Since $c\in C^0$ but $c \notin H^1,$ the antiderivative $f(x) = \int_0^x c$ is $C^1$ but not $H^2$.
In general, you can always adapt this idea by taking a product and localizing, i.e. letting $f(x,y,\ldots,z) = \eta(x,y,\ldots,z)\int_0^x c$ where $\eta$ is some smooth bump function supported in $\Omega.$