Here is an example of a connected Lie group acting on $M$, such that the action does not extend to $W$ but extends to some $W'$ with $\partial W'=M$. All the existing actions are analytic isometric, and all obstructions are continuous, so the problem is not regularity.

Take $G=M=\mathbb T^1$ the 1-torus, acting on itself in the obvious fashion, by translations. Clearly, if $M$ is seen as the boundary of the disc $W'=\mathbb D^2$, then the action on $M$ extends to an action on $W'$, again by rotations. (In the following I prefer to think of $\mathbb T^1$ as $\mathbb R/\mathbb Z$.)

However, let $W$ be a 2-torus with a disc removed, and identify $M$ with the boundary of $W$. Assume by contradiction that the induced action on $\partial W$ is the restriction of an action on $W$.

Let $Z$ be the manifold constructed as the gluing of $[0,1]\times M$ with $W$ along $\lbrace1\rbrace\times M$ and $\partial W$. Define an action of $\mathbb R$ on $Z$ in such a way that the boundary of $Z$ (which corresponds to $\lbrace0\rbrace\times M$) is fixed, and the restriction to $W$ is the action induced by that of $\mathbb T^1$ on $W$. On $[0,1]\times M$, we have to continuously Dehn-twist the collar to match the rotation of $\lbrace1\rbrace\times M$.

Now the action $f$ at time $1$ is homotopic to the identity of $Z$ relative to its boundary (in the strong sense), but $f$ is just a Dehn twist of $[0,1]\times M$, since $f$ is the identity in $W$ (1 has image 0 in $\mathbb T^1$). It is probably classical that this Dehn twist is not homotopic to the identity relatively to the boundary, personally I convinced myself on a picture that the induced map $f_*$ on $\pi_1(M,x)$ for $x$ on the boundary is not the identity; in any case, this is a contradiction.

2more comments