I think one does need to be careful (c.f. the comments). There are Radon measures for which locally null and null are different. The canonical example is to take $X=\mathbb R^2$ with the topology that $U$ is open if and only if $U_x = \{ y : (x,y)\in U \}$ is open, for each $x$. Then define a Radon measure by the functional $$ C_{00}(X)\rightarrow\mathbb C; \quad f\mapsto \sum_x \int f(x,y) \ dy $$ where we use Lebesgue measure. Then $\{(x,0)\}$ is locally null, but not null. See exercises 3.3.6 and 7.2.4 in Cohn's book.

I think the resolution to the problem asked is to realise that locally compact groups are not entirely arbitrary locally compact spaces. The "trick" which Folland uses in his Harmonic Analysis book is to realise that for any $G$ we can always find an open and closed, $\sigma$-compact subgroup $H$. Then $G/H$ has the discrete topology, and being $\sigma$-compact, everything works fine on $H$. Then you can reconstruct the results you want by working piecewise on each coset of $H$. (See page 51 of Folland).

This resolution works quite adequately with "construction 2". Indeed, see Cohn, Theorem 9.4.8 which shows exactly that $L^\infty$ is the dual of $L^1$ for any regular Borel measure on $G$. However, beware of the subtle point in the proof, that you cannot exactly just work on each coset of $H$: to get a Borel function, some trick using Lusin's theorem (to approximate by continuous functions) is needed.

(An alternative approach to "fix" the duality issue is to work with a larger $\sigma$-algebra than the Borel sets; compare Cohn Section 7.5. If I understand things right, this is the same notion of "measurability" which Hewitt+Ross uses. Here, Exercise 7.5.5 in Cohn is interesting: $L^p$ for this $\sigma$-algebra, or $L^p$ for Borel sets, are isometrically isomorphic, so long as $p$ is finite.)

In conclusion, the two constructions asked about are in general different. But in this special case (locally compact groups) they agree.

I think one does need to be careful (c.f. the comments). There are Radon measures for which locally null and null are different. The canonical example is to take $X=\mathbb R^2$ with the topology that $U$ is open if and only if $U_x = \{ y : (x,y)\in U \}$ is open, for each $x$. Then define a Radon measure by the functional $$ C_{00}(X)\rightarrow\mathbb C; \quad f\mapsto \sum_x \int f(x,y) \ dy $$ where we use Lebesgue measure. Then $\{(x,0)\}$ is locally null, but not null. See exercises 3.3.6 and 7.2.4 in Cohn's book.

I think the resolution to the problem asked is to realise that locally compact groups are not entirely arbitrary locally compact spaces. The "trick" which Folland uses in his Harmonic Analysis book is to realise that for any $G$ we can always find an open and closed, $\sigma$-compact subgroup $H$. Then $G/H$ has the discrete topology, and being $\sigma$-compact, everything works fine on $H$. Then you can reconstruct the results you want by working piecewise on each coset of $H$. (See page 51 of Folland).

This resolution works quite adequately with "construction 2". Indeed, see Cohn, Theorem 9.4.8 which shows exactly that $L^\infty$ is the dual of $L^1$ for any regular Borel measure on $G$. However, beware of the subtle point in the proof, that you cannot exactly just work on each coset of $H$: to get a Borel function, some trick using Lusin's theorem (to approximate by continuous functions) is needed.

(An alternative approach to "fix" the duality issue is to work with a larger $\sigma$-algebra than the Borel sets; compare Cohn Section 7.5. If I understand things right, this is the same notion of "measurability" which Hewitt+Ross uses. Here, Exercise 7.5.5 in Cohn is interesting: $L^p$ for this $\sigma$-algebra, or $L^p$ for Borel sets, are isometrically isomorphic, so long as $p$ is finite.)

In conclusion, the two constructions asked about are in general different. But in this special case (locally compact groups) they agree.

A meta-question is: which construction to use? I guess I don't know (and it doesn't matter, if you believe this answer!) From my experience, the literature in abstract harmonic analysis almost always uses construction 1, and the "locally" language. Furthermore, in most cases, it doesn't really matter exactly which $\sigma$-algebra one works with.

More pragmatically, the vast majority of examples are $\sigma$-compact, and so there is relatively little to be lost by just assuming that $G$ is $\sigma$-compact. Then it's clear that both constructions agree. (Related to the often seen "In this paper we assume that all Hilbert spaces are separable" caveats).