I think the question mainly highlight either of two misunderstandings:
- misunderstanding about the definition of heritability
- misunderstanding why the slope of the regression line is equal to heritability (in the narrow sense).
Definition of heritability
Please have a look at the post Why is a heritability coefficient not an index of “how genetic” something is?.
Notation
Please note that, here I use $h^2 = \frac{V_A}{V_P}$ to designate the heritability in the narrow sense, where $V_A$ is the additive genetic variance and $V_P$ is the (total) phenotypic variance. If anything of what I just said is unclear, please refer to the aforementioned post.
Without any loss of generalization, we will set our scale so that the average phenotype in any studied case is equal to zero. From this starting value, we note X (with a subscript) the contribution of a given haplotype (or allele) to deviate the phenotype of an individual. We will note $E$ (with a subscript) the contribution of a given environmental state to deviate the phenotype of an individual.
An individual phenotype
Let's consider an individual O (stands for offspring, it will makes sense later). The phenotype of this individual is given by the addition of the contribution of each haplotype and of the environment. The expected phenotype of O (called $P_O$) is therefore
$$P_O=X_m + X_p + e_O$$
, where $X_m$ is the contribution of the maternal haplotype, $X_p$ is the contribution of the paternal haplotype and $e_O$ is the contribution of the environment.
Just a note
It might be worth skipping this note and coming back to it once you understood the below calculations (but I did not want to move this note to the end).
In the above calculation, one must assume absence of epistasis and dominance effect. I am not exactly sure how this might eventually cancel out later or whether results would then differ. I would welcome a follow-up in the comments to clarify this point but it might be worth its own post. As the question refers to the heritability of intelligence (assuming this term is defined), I would like to highlight that it is unlikely that we have much understanding about the exact pattern of epistasis underlying this trait as measure of epistatic interaction are often difficult to achieve.
Phenotype of O's parent
Let's note $P_P$, the phenotype of one parent of O'. The below notation applies to the mother but calculations would be the same for the father.
$$P_P = X_m + X_m' + e_P$$
, where $X_m'$ is the haplotype of the mother that has NOT been transmitted to the offspring and $e_P$ is the contribution of the environment experienced by the parent (mother).
Covariance between $P_O$ and $P_P$
The covariance between $P_O$ and $P_P$ (noted $cov(P_O, P_P)$ is just the sum of the covariances of all contributions, that is
$$cov(P_O,P_P) = cov(X_m,X_m) + cov(X_m,X_m') + cov(X_m,e_m) + cov(X_P,X_m) +\
cov(X_P,X_m') + cov(X_P,e_m) + cov(e_O,X_m) + cov(e_O,X_m') + cov(e_O,e_m)$$
I let you take a few minutes to think about the values of each of these covariances.... They are almost all equal to zero (are independent). There are only two covariances that (under a simple model) can potentially be different from 0; $cov(X_m,X_m)$ and $cov(e_O,e_m)$. $cov(e_O,e_m)$ is teh correlation between the environment experienced by the mother and the one experienced by the father. For simplicity, we will make the assumption that these two environments are independent and therefore $cov(e_O,e_m)=0$. The above, long and hairy equation therefore reduces to
$$cov(P_O,P_P) = cov(X_m,X_m) = var(X_m) = \frac{V_A}{2}$$
Correlation between $P_O$ and $P_P$
By definition the correlation (which is the slope of the regression line) is just the covariance divided by the total phenotypic variance. Therefore the correlation between $P_O$ and $P_P$ (noted $cor(P_O,P_P)$) is
$$cor(P_O,P_P) = \frac{V_A}{2} \frac{1}{V_P} = \frac{h^2}{2}$$
Generalization to any correlation
Following the above logic, one can show that in general the correlation between any two individual phenotypes $P_X$ and $P_Y$ is
$$cor(P_X,P_Y) = r h^2$$
, where $r$ is the coefficient of relationship. For parent-offspring $r=\frac{1}{2}$.
Particular case of mid-parent vs offspring relationship
This is just a study case to show that when consdiering a made-up phenotype such as a hypoethetical mid-parent phenotype, then the math are just a little bit more tricky. If you consider the mid-parent phenotype $P_{mid}$, then you must consider that the total phenotypic variance is half the actual phenotypic variance. The correlation therefore becomes
$$cor(P_{mid},P_O) = \frac{cov(P_{mid},P_O)}{\sqrt{\frac{V_P^2}{2}}} = \frac{h^2}{\sqrt{2}}$$
Conclusion
All of the above mean that if you plot the phenotypes of a series of individual pairs, where each pair is related by $r$, then the slope of the regression line is $r h^2$. For example, if you plot the phenotype of the mothers with their offspring, then the slope of the regression is equal to $\frac{h^2}{2}$. If $h^2=0.4$ (which is a typical value of heritability), then the mother-offspring regression will have a slope of $0.2$.
