The situation is indeed a delicate one and one needs to carefully check the conventions before transferring a result from one context to another. The situation is summarized in Royden's Real Analysis, though with just a few examples.
For a given space $X$, the main players are:
$\mathcal{B}a$ — the σ-algebra generated by the compact Gδ sets. (The Baire algebra.)
$\mathcal{B}c$ — the smallest σ-algebra with respect to which all continuous real-valued functions are measurable.
$\mathcal{B}k$ — the σ-algebra generated by the compact sets.
$\mathcal{B}o$ — the σ-algebra generated by the closed sets. (The Borel algebra.)
$\mathcal{S}$ — the smallest σ-ring containing the compact sets.
$\mathcal{R}$ — the smallest σ-ring containing the compact Gδ sets.
In general, we have the inclusions $$\mathcal{B}a \subseteq \mathcal{B}k \subseteq \mathcal{B}o\quad\mbox{and}\quad\mathcal{B}a \subseteq \mathcal{B}c \subseteq \mathcal{B}o,$$ but $\mathcal{B}c$ and $\mathcal{B}k$ are not necessarily related. Moreover, $\mathcal{B}a = \mathcal{B}c \cap \mathcal{B}k$ and $\mathcal{B}o$ is generated by $\mathcal{B}c \cup \mathcal{B}k$. The σ-rings $\mathcal{S}$ and $\mathcal{R}$ consist of the σ-bounded elements of $\mathcal{B}a$ and $\mathcal{B}o$, respectively.
When $X$ is σ-compact and locally compact, then $$\mathcal{R} = \mathcal{B}a = \mathcal{B}k mathcal{B}c \quad\mbox{and}\quad\mathcal{S} = \mathcal{B}c mathcal{B}k = \mathcal{B}o.$$ A compact example showing the strict inequality is $\beta\mathbb{N}$.
When $X$ is metrizable, more generally when closed sets are Gδ, we have $$\mathcal{R} = \mathcal{S} \subseteq \mathcal{B}a = \mathcal{B}k \subseteq \mathcal{B}c = \mathcal{B}o.$$ An example showing strict inequality is the space of irrational numbers.
When $X$ is locally compact and separable, then we have $$\mathcal{R} = \mathcal{S} = \mathcal{B}a = \mathcal{B}k = \mathcal{B}c = \mathcal{B}o.$$
That said, the uniqueness (up to normalization) of Haar measure and it's (inner) regularity guarantee that all constructions will agree on their common domain of definition. Proving this from scratch does require some work depending on where you start and end.
