Q1: Definately not always. More like "almost never". If the automorphism extends to $\mathbb R^3$, then the bundle $S^1 \ltimes_f M$ would embed in $S^4$. $S^1 \ltimes_f M$ is the bundle over $S^1$ with fiber $M$ and monodromy $f$. The most-commonly used obstructions to embedding in this case are things like the Alexander polynomial, and Milnor signatures.
I don't see where the metric on $M$ plays a role for this.
If you want to see automorphisms that extend (and do not extend) for your Q1, take a look at my arXiv paper. You'll also find some references to several Jonathan Hillman papers that explore such obstructions.
In the case that your surface is unknotted -- bounding handlebodies on both sides (thinking of $M \subset S^3$) then the automorphisms of $M$ that extend in this case are well-known. They're called the mapping class group of a Heegaard splitting of $S^3$. It's an infinite group. Generators are known for it (if I recall, they're the automorphisms induced by handle slides) but off the top of my head I'm not sure how much is known about the structure of the group. Do a little Googling on "mapping class group of a Heegaard splitting of S^3" and you should start finding relevant material.
To respond to your 2nd edit, if the co-dimension is high enough all automorphisms extend. This is a theorem of Hassler Whitney's. The basic idea is this, let $f : M \to M$ be an automorphism. Let $i : M \to \mathbb R^k$ be any embedding. So you have two embeddings $i \circ f$ and $i$ of $M$ in $\mathbb R^k$. Any two maps $M \to \mathbb R^k$ are isotopic provided the co-dimension is large enough $k \geq 2m+3$ suffices, for example. So isotope your standard inclusion to the one pre-composed with $f$. The Isotopy Extension Theorem gives you the result.
For example, if $\Sigma$ is a Heegaard splitting / the surface is unknotted, $\Sigma \subset \mathbb R^3$ (or $\subset \mathbb S^3$) and you have an automorphism $f : \Sigma \to \Sigma$ a neccessary and sufficient condition for $f$ to extend to $\mathbb R^3$ (or a side-preserving automorphism of the pair $(S^3,\Sigma)$ in the $S^3$ case) is that if $C \subset \Sigma$ is a curve that bounds a disc on either the inside or outside of $\Sigma$ respectively, then $f(C)$ bounds a disc on the inside or outside of $\Sigma$ respectively (here I'm using inside/outside re the Jordan-Brouwer separation theorem). Since the fundamental group of the complement is just a free product of infinite cyclic groups this is something that can be checked rather easily provided you know the map $f$ well enough.