The question of the non-relativistic limit is obscured by the fact that the non-relativistic kinematic symmetry group, as we now understand it, has 11 dimensions, with the inclusion of the eleventh generator - the central charge - which means that we're actually talking about the Bargmann group, not the Galilei group, while the kinematic symmetry group for relativity has only 10 dimensions, because historically the symmetry transforms were devised before we had our current understanding of the situation in non-relativistic theory.
The formulation of Relativistic symmetry was never retro-updated to match this updated view of non-relativistic symmetry, so as to retain the validity of the correspondence limit. The correspondence limit was broken by the revision of non-relativistic theory.
In order to have consistency with the correspondence limit, you also need an 11th generator for the Poincaré group, and adding in the trivial central extension will also happen to clarify a great number of other issues.
Everything centers around the issue of the mass and energy. I make a distinction between the kinetic energy used in non-relativistic theory, denoting it $H$, versus the "total" energy used in relativity, denoting it $E$; while also using the "moving mass" $M$ interchangeably, where $E = M c^2$. The total energy $E$ can be decomposed into the kinetic part $H$ plus an invariant part $μc^2$ proportional to a mass invariant $μ$. That's the additional generator. The use of $M$ is preferable, since it has a meaningful non-relativistic limit, by virtue of the relation $M = μ + H/c^2$, while $E$ does not.
In Weinberg's treatment (in section 2.4), he actually does make a similar distinction - his $W$ corresponds to my $H$, and his $H$ to my $E$, and his $M$ to my $μ c^2$, so you could write $E = μ c^2 + H$, in place of his $H = M + W$. So, he sneaked in the 11th generator too, but tried to hand-way his way out of explicitly calling it out as such.
This is a slight generalization of the Poincaré symmetry of Relativity that reduces to Relativity - at least for Tardyons - by identifying $μ$ with the rest mass $m$. In place of the invariant $m^2 c^2 + P^2 - E^2/c^2$, where $𝐏$ is the momentum 3-vector, you have the quadratic invariant $λ = P^2 - 2MH + H^2/c^2$, whose non-relativistic limit is just $P^2 - 2MH$. For the linear invariant $μ = M - H/c^2$, the non-relativistic limit is just $μ = M$. The restriction of the generalized Poincaré symmetry to Relativity then corresponds to setting $λ = 0$.
Otherwise, if you allow for a non-zero rest value for $H$, as an invariant $U$, then (for Tardyons) in the rest state (where $P^2 = 0$ and $M = m$, the rest mass), you would have $λ = -2mU + U^2/c^2$ and $μ = m - U/c^2$. In the non-relativistic limit the quadratic invariant yields the relation $P^2 - 2MH = -2mU$, with $M = m = μ$ all coinciding, and the decomposition $H = P^2/2m + U$: the kinetic energy for the center of mass motion plus the internal energy. So, the generalization allows you extra room to add in an internal energy. In the general case, the rest mass would only be defined by
$$m^2 = μ^2 - \frac{λ}{c^2} = M^2 - \frac{P^2}{c^2},$$
and only for Tardyons, where $M^2 > P^2/c^2$. It is an abuse of terminology to use the term "rest mass" with Luxons or Tachyons, when they have no rest frame! At best, you can catch Luxons on the margin by pretending $m = 0$; when actually all you're really saying is that $P = M c$, or equivalently: $E = P c$. But, there's no such thing as a "rest mass" for Luxons, since they can never be at rest.
Once you do all of that, then the situation with the $[K,P]$ brackets becomes clear. In indexed form, they would be written as
$$[K_i, P_j] = iħ δ_{ij} M = iħ δ_{ij} \left(μ + \frac{H}{c^2}\right) = iħ δ_{ij} \frac{E}{c^2},$$
and its non-relativistic limit would be $[K_i, P_j] = iħ δ_{ij} μ$, as expected, with $μ$ as the central charge.
As to what the transforms correspond to geometrically, you're not going to find any answers in 4-dimensional geometry at all. You can already see that, non-relativistically, by the fact that the mass, energy and momentum transform together as a 5-vector, not as a 4-vector. So, for consistency, you also need a 5th coordinate to serve as conjugate to the mass, with the invariant 1-form (from which the operator correspondence is ultimately drawn) written down as:
$$𝐏·d𝐫 - H dt + μ du, \hspace 1em 𝐫 = (x, y, z).$$
The action under an infinitesimal boost by $𝞄$ is then given by
$$ΔH = -𝞄·𝐏, \hspace 1em Δ𝐏 = -𝞄M, \hspace 1em ΔM = -α𝞄·𝐏,$$
with the non-relativistic case ($α = 0$) and relativistic case ($α = 1/c^2 > 0$) being distinguished from one another by the deformation parameter $α$.
The (trivial) central extension of the Poincaré group is a deformation of the centrally-extended Galilei group, which is called the Bargmann group.
As a footnote: it's possible to embed these two groups into a much larger family of kinematic symmetry groups (that includes, for instance the [anti-]de Sitter groups), as a three parameter family of deformations of the "static group" ... and to provide a uniform 5-dimensional geometric representation applies across the board to the central extensions of them all.
The coordinate transforms which make the above one-form invariant are given in infinitesimal form as
$$Δ𝐫 = -𝞄t, \hspace 1em Δt = -α𝞄·𝐫, \hspace 1em Δu = 𝞄·𝐫,$$
and yield the following two geometric invariants:
$$|d𝐫|^2 + 2 dt du + α du^2, \hspace 1em ds ≡ dt + α du.$$
Minkowski geometry is recovered by setting the quadratic invariant to 0 and identifying $s$ as the proper time. So, both the relativistic and non-relativistic geometries are contained in this geometry as light cones. The ambient (4+1)-dimensional geometry, in the case $α = 0$ is called the Bargmann geometry. It is still (4+1)-dimensional, for all other values of $α$, the only difference being on the direction and nature of the invariant $ds$ within this geometry.
If you integrate the infinitesimal transforms to finite form, taking boosts in the $x$ direction, the resulting transforms are
$$x → \frac{x - vt}{\sqrt{1 - α v^2}}, \hspace 1em y → y, \hspace 1em z → z, \hspace 1em t → \frac{t - αvx}{\sqrt{1 - α v^2}},\\ \hspace 1em u → u + \frac{v x}{\sqrt{1 - α v^2}} - \frac{v^2}{1 + \sqrt{1 - α v^2}} \frac{t}{\sqrt{1 - α v^2}}.$$
The coordinate $u = c^2 (s - t)$ is just the time-dilation, itself, ramped up by an order of $c^2$. Ramping it up gives it a non-zero and non-trivial non-relativistic limit, in contrast to $s - t$ which goes to zero. So, the extra coordinate is a non-relativistic vestige of time-dilation.
In the non-relativistic limit, this transform becomes:
$$x → x - vt, \hspace 1em y → y, \hspace 1em z → z, \hspace 1em t → t,\hspace 1em u → u + v x - \frac{v^2}{2} t.$$
So, now do the translation and boosts. Translation first, then boost:
$$x → x + a → x + a - vt, \hspace 1em y → y → y, \hspace 1em z → z → z, \hspace 1em t → t → t, \\ u → u → u + v (x + a) - \frac{v^2}{2} t.$$
Boosts first, then translation:
$$x → x - vt → x - vt + a, \hspace 1em y → y → y, \hspace 1em z → z → z, \hspace 1em t → t → t, \\ u → u + v x - \frac{v^2}{2} t → u + v x - \frac{v^2}{2} t.$$
As you can see: they no longer commute. If you do the first, followed by the inverse of the second, the net result is a translation in $u$; and - as noted above - the mass invariant $μ$ has $u$ as its conjugate, so it corresponds to $u$ translation.
If you look carefully at Weinberg's treatment, you'll be able to spot where he slipped in vestiges of the $u$ coordinate. It sneaked into his treatment (at least in vestigial form), even beneath his own guard.
Here's a hint. The non-relativistic Schrödinger equation has a differential operator that is the operator version of $\hat{H} - \hat{P}^2/2m$. It should actually be written as $\hat{P}^2 - 2\hat{M}\hat{H}$, with an extra operator eigenvalue equation for $\hat{μ} = m$, if restricting attention to Tardyons. The solutions, when applying this to $ψ$, when expressed as a function of $u$, are given by $ψ(u) = e^{imu/ħ} ψ(0)$. If you assume that the operator $\hat{μ} = -iħ ∂/∂u$ indexes "superselection" sectors (as Weinberg did); i.e. that it commutes with all observables, then that actually expresses the $u$-independence of the observables: a kind of gauge symmetry.