= Solution
Let $f=-V^{-1}\log\mathcal Z$ be the dimensionless <free-energy density>. The transformed volume is $V'=V/B$. Including the normalization from eliminated modes gives
$$
f(t,h)=g_b(t,h)+B^{-1}f(t',h'),
$$
where $g_b$ is an additive shell contribution. After subtraction of regular backgrounds, the leading singular part obeys the <scaling hypothesis for critical phenomena>
$$
f_s(t,h)=b^{-D_{\rm eff}}f_s(b^2t,b^{(d+5)/4}h),\qquad D_{\rm eff}=(d+1)/2.
$$
Choose $b=t^{-1/2}$ for the high-temperature side $t>0$. Then the nominal homogeneous form is
$$
\boxed{f_s(t,h)=t^{(d+1)/4}g_f\!\left(h/t^{(d+5)/8}\right),\quad
\alpha=(7-d)/4,\quad\Delta=(d+5)/8.}
$$
The volume exponent is $D_{\rm eff}$, so inserting $d$ in the isotropic hyperscaling formula would give the wrong answer.
There are two qualifications to interpreting this as the total <free energy>. Completing the Gaussian square gives the source contribution exactly as $-h^2/(2t)$, and the zero-field <determinant> contributes
$$
f(t,h)=\frac12\int_q\log(Kq_\parallel^2+Lq_\perp^4+t)-\frac{h^2}{2t}+\text{normalization}.
$$
Rescaling $q_\parallel\sim t^{1/2}$ and $q_\perp\sim t^{1/4}$ gives the power $t^{(d+1)/4}$, while $2\Delta-1=(d+1)/4$ makes the source term consistent with it. Cutoff-dependent analytic terms need not obey the homogeneous law. Furthermore <Gaussian free-energy logarithms at integer singular powers> replace the pure zero-field power by $t^n\log t$ when $(d+1)/4=n$ is a positive integer, including $d=3,7$. This is the usual logarithmic qualification of the nominal exponents. The quadratic model also has no stable negative-$t$ phase unless a stabilizing interaction is added.
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