Yes, the two are the same, but I'd be inclined to separate the issues into first converting automorphic forms (with automorphy factors) on the domain $\mathfrak H$ into automorphic forms on the group $G=SL_2(\mathbb R)$ or $GL_2(\mathbb R)$, and look at the measures as a separate issue.

That is, for $f(\gamma\cdot z)=f(z)\cdot j(\gamma,z)$ for some cocycle $j$ that extends to a cocycle on $G\times \mathfrak H$, let $F_f(g)=f(g\cdot i)\cdot j(g,i)^{-1}$. This is now left $\Gamma$-invariant (no cycle) and right $K=SO(2,\mathbb R)$-equivariant by $g\to j(g,i)^{\pm 1}$. The Petersson inner product $\langle f_1,f_2\rangle$ is the integral of $f_1(z)\bar{f}_2(z)y^{2k}$, and the latter is $\Gamma$-invariant. Further, this is equal to $F_{f_1}(g)\cdot\bar{F}_{f_2}(g)$ for any $g\in G$ such that $g(i)=z$.

The measures on all the available physical spaces are uniquely determined from the Haar measures on the various groups $GL_2(\mathbb R)$ and $GL_2(\mathbb Q_p)$. In fact, if we follow Weil's prescription (as in "Adeles and Algebraic Groups") we see that the product of all the local Haar measures is uniquely determined up to a constant. (Cf. "Tamagawa measure".)

Whether or not we use that prescription, the only thing needed to see that the various measure determine each other is that very general property that for any topological group $G$ and closed subgroup $H$, $G/H$ admits a left $G$-invariant measure if and only if the modular function of $G$ restricted to $H$ is the modular function of $H$, and, if so, then $$ \int_G f(g)\,dg = \int_{G/H} \int_H f(gh)\;dh\;dg $$ for $f\in C^o_c(G)$.

In the case at hand, the left $G$-invariant measure on $\mathfrak H\approx G/K$ is thus determined by choice of total measure on $K$ and choice of Haar measure on $G$.

To subsequently make automorphic forms on the Lie group $G$ into automorphic forms on the adele group, some sort of Strong Approximation suffices, but we still need to specify the constant multiple of Haar measure we take on $SL_2(\mathbb Q_p)$, etc.

Sure, we can specify the Haar measure on $SL_2(\mathbb A)$ or $GL_2(\mathbb A)$ all at once, too, rather than factor-wise.

In any case, I think there is no need to work so hard as to talk about measures on quotients of spaces by groups. All that is needed is quotients $G/H$ (or $H\backslash G$).

Yes, the two are the same, but I'd be inclined to separate the issues into first converting automorphic forms (with automorphy factors) on the domain $\mathfrak H$ into automorphic forms on the group $G=SL_2(\mathbb R)$ or $GL_2(\mathbb R)$, and look at the measures as a separate issue.

That is, for $f(\gamma\cdot z)=f(z)\cdot j(\gamma,z)$ for some cocycle $j$ that extends to a cocycle on $G\times \mathfrak H$, let $F_f(g)=f(g\cdot i)\cdot j(g,i)^{-1}$. This is now left $\Gamma$-invariant (no cycle) and right $K=SO(2,\mathbb R)$-equivariant by $g\to j(g,i)^{\pm 1}$. The Petersson inner product $\langle f_1,f_2\rangle$ is the integral of $f_1(z)\bar{f}_2(z)y^{2k}$, and the latter is $\Gamma$-invariant. Further, this is equal to $F_{f_1}(g)\cdot\bar{F}_{f_2}(g)$ for any $g\in G$ such that $g(i)=z$.

The measures on all the available physical spaces are uniquely determined from the Haar measures on the various groups $GL_2(\mathbb R)$ and $GL_2(\mathbb Q_p)$. In fact, if we follow Weil's prescription (as in "Adeles and Algebraic Groups") we see that the product of all the local Haar measures is uniquely determined up to a constant. (Cf. "Tamagawa measure".)

Whether or not we use that prescription, the only thing needed to see that the various measure determine each other is that very general property that for any topological group $G$ and closed subgroup $H$, $G/H$ admits a left $G$-invariant measure if and only if the modular function of $G$ restricted to $H$ is the modular function of $H$, and, if so, then $$ \int_G f(g)\,dg = \int_{G/H} \int_H f(gh)\;dh\;dg $$ for $f\in C^o_c(G)$.

In the case at hand, the left $G$-invariant measure on $\mathfrak H\approx G/K$ is thus determined by choice of total measure on $K$ and choice of Haar measure on $G$.

To subsequently make automorphic forms on the Lie group $G$ into automorphic forms on the adele group, some sort of Strong Approximation suffices, but we still need to specify the constant multiple of Haar measure we take on $SL_2(\mathbb Q_p)$, etc.

Sure, we can specify the Haar measure on $SL_2(\mathbb A)$ or $GL_2(\mathbb A)$ all at once, too, rather than factor-wise.

In any case, I think there is no need to work so hard as to talk about measures on quotients of spaces by groups. All that is needed is quotients $G/H$ (or $H\backslash G$).

Edit: left $\Gamma$-invariant right $K$-invariant functions on $G$ are in $\Gamma\backslash H\approx \Gamma\backslash G/K$, but/and the latter is conveniently the right $K$-fixed elements in functions on $\Gamma\backslash G$, e.g., $L^2(\Gamma\backslash G)^K$. Since $K$ is compact, all the useful relations hold... In particular, the measure on $\Gamma\backslash G$ is the natural right $G$-invariant one, and the right action of $G$ is unitary, etc. That is, removing the obstacle of right-quotient by $K$ is just a matter of viewpoint.

Ignoring the center for notational simplicity, strong approximation gives $G_k\cdot G_\infty \cdot K_f=G_{\mathbb A}$ for any compact-open in the finite component. Thus, $$(G_k\cap K_f)\backslash G_\infty \approx G_k\backslash G_{\mathbb A}/K_f \approx (G_k\backslash G_{\mathbb A})^{K_f}$$ That is, the automorphic forms of various levels are images by the projectors associated to $K_f$ on $L^2(G_k\backslash G_{\mathbb A})$.