The first theorem you're talking about is known as Dini's theorem. I know the proof for $f:[a,b]\to \Bbb R$, but it can be adjusted for your case.
THEOREM Suppose $\{f_n\}$ is a sequence of continuous functions from $[a,b]$ to $\Bbb R$ that converge pointwise to a continuous function $f$ over $[a,b]$. If $f_{n+1}\leq f_n$, then convergence is uniform.
PROOF We set $g_n=f_n-f$ and note that $g_n\geq g_{n+1}$ and the $g_n$ are continuous, converging pointwise to $0$. For a given $\epsilon>0$. Consdier the (relatively) open sets (because of continuity of the $g_n$) $O_n=\{x\in [a,b] :g(x)<\epsilon\}$. Note that since $g_n\geq g_{n+1}$ we have $O_n\subset O_{n+1}$. Given $x\in[a,b]$ there is an $n$ such that $g_n(x)<\epsilon$; whence $\bigcup_{n\in\Bbb N}O_n=[a,b]$. But since $[a,b]$ is compact there exists a finite set $K=\{1,\dots,m\}$ such that $\bigcup_{k=1}^m O_{n_k}=[a,b]$. But since $O_n\subset O_{n+1}$ the greatest element of $K$, call it $\ell$, is such that $O_\ell =[a,b]$. And we're done: for every $n\geq \ell$ we have $g_n(x)<\epsilon$; as desired. $\blacktriangle$
ADD A little googling brought up this, which is where I learned the proof from. It has a few examples showing the hypotheses
$(1)$ $f_n$ decreases with $n$
$(2)$ The domain is compact.
$(3)$ $f$ is continuous.
are all essential.
Best Answer
Partial answer:
If $y=f(x)$ and $y_n=f_n(x)$ then $$ |g(y)-g_n(y)|=|x-g_n(\;y-y_n+y_n\;)|=|g_n(y_n)-g_n(y_n + e_n)| $$ where $e_n=y-y_n \rightarrow 0$.
Therefore if $\{g_n\}$ is uniformly equicontinuous the answer is yes: $$ |g(y)-g_n(y)|\le \sigma(e_n) $$ where $\sigma$ is the modulus of continuity of the family $\{g_n\}$.