I have never been trained in the art of differential geometry and I really have no one else to turn to, so if someone can kindly give me some guidance on the following questions, I would really appreciate it. I really just need a yes or no answer, mainly to make sure I'm not on the wrong path.
Let $U \subset R^d$ be a connected, bounded open subset, suppose $f:U\to R^d$ is infinitely differentiable.
Is it true that we can find a sequence $(f_n)_{n\geq 1} \subset C^\infty_c(R^d)$ (compactly supported) and $(D_n)_{n\geq 1}$ such that $D_{n-1} \subset \overline{D_{n-1}} \subset D_n \subset \overline{D_n} \subset U$, $\bigcup_{n\geq 1} D_n = U$, $f_n = f$ on $D_n$, $f_n$ has support in $D_{n+1}$?
Suppose $f \in C^\infty_c(R^d)$, $D \subset R^d$ a non-empty, connected, bounded open subset and $f = 0$ on $\partial D$, then is the function $x \in D \mapsto \frac{|f(x)|}{d(x,\partial D)}$ bounded above?
Yes, both of these are true.
For your first question, you can get the sets $D_n$ by shrinking $U$ by $1/n$; i.e. let $D_n = \{ x \in U : \overline{B(x,1/n)} \subset U \}$. You should be able to verify that the $D_n$ are open and satisfy your requirements. Then just define $f_n$ by multiplying $f$ by a bump function $\eta_n$ supported on $D_{n+1}$ satisfying $\eta_n|_{D_n} = 1$. The existence of such a function is known as the Smooth Urysohn Lemma if you want to look up a proof.
For your second question, we can get an explicit bound in terms of the derivative of $f$. Since $\partial D$ is closed and bounded, it is compact and thus for each $x\in D$ there is a $y \in \partial D$ with $|x-y| = d(x,\partial D)$. Using $f(y)=0$, the fundamental theorem of calculus, the chain rule and the Cauchy-Schwarz inequality, we can estimate
$$|f(x)| = |f(x) - f(y)| = \left|\int_0^1 \frac d{dt} f(tx + (1-t)y)dt\right| \\=\left|\int_0^1 (x-y)\cdot\nabla f(tx + (1-t)y)dt\right| \le|x-y| \sup |\nabla f|.$$
Since $|x-y| = d(x,\partial D)$, this implies $|f(x)|/d(x,\partial D) \le \sup |\nabla f|$, which is independent of $x$. This bound is finite because $|\nabla f|$ is a continuous compactly supported function.