Well, one trivial connection is that if you look at $1\times 1$ matrices (which have only a single complex entry), then you'll find that it is real iff it is Hermitian, its complex conjugate is its conjugate transpose, and its polar decomposition is the polar form.
Also, just like a complex number can be uniquely decomposed into a real and an imaginary part ($z = a+\mathrm ib$ with real $a,b$), a complex matrix can be uniquely decomposed into a Hermitian and an "anti-Hermitian" part, i.e,. $M =A + \mathrm iB$ with $A$ and $B$ Hermitian. And just like $\Re(z)=\frac12(z+\bar z)$ and $\Im(z)=\frac1{2\mathrm i}(z-\bar z)$, the Hermitian part of a matrix is $\frac12(M+M^*)$ and the "anti-Hermitian" part is $\frac1{2\mathrm i}(M-M^*)$.
Moreover, just like $\bar zz$ is a non-negative real number, $M^*M$ is a positive semidefinite matrix.
Another point: Hermitian matrices have real eigenvalues, and unitary matrices have eigenvalues of the form $\mathrm e^{\mathrm i\phi}$.
About the usefulness of the analogy:
In classical physics, observables should be real. In quantum physics, observables are represented by Hermitian matrices. Also, the quantum analogue to probability densities, which are non-negative functions with integral $1$, are density operators, which are positive semidefinite matrices with trace $1$. So there's indeed some connection.
The matrix $uu^T$ is a special case of $\mathsf{PSD}$, (Positive semidefinite), matrices called a symmetric dyad.
Indeed, if $A = uu^T$ for some vector $u \in \mathbf{R}^n$, then for every $x$:
$$q_A (x) = x^T uu^T x = (u^Tx)^2 \ge 0. $$
Another way to look at this is thus:
$$\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \cdot \begin{bmatrix} x_1 & x_2 & \ldots & x_n \end{bmatrix} = \begin{bmatrix}x_1^2 & x_1x_2 & \ldots & x_1x_n \\ x_2x_1 & x_2^2 & \ldots & x_2x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_nx_1 & x_nx_2 &\ldots &x_n^2 \end{bmatrix}$$
What can you say about the above matrix? In particular, it's symmetric, and what property do all of its leading principal minors have in common?
To answer your first question, I would direct you to an answer of Plainview's in
the following question: Is the product of symmetric positive semidefinite matrices positive definite?
The product of two symmetric $\mathsf{PSD}$ matrices is $\mathsf{PSD}$, iff the product is also symmetric. More generally, if $A$ and $B$ are $\mathsf{PSD}$, $AB$ is $\mathsf{PSD}$ iff AB is normal, ie, $(AB)^T AB = AB(AB)^T$.
Since your matrix $H$ is positive definite, it is also positive semidefinite. Therefore, the product $HD$ is positive semidefinite iff
$$
(HD)^T HD = HD(HD)^T
$$
Best Answer
One intuitive definition is as follows. Multiply any vector with a positive semi-definite matrix. The angle between the original vector and the resultant vector will always be less than or equal $\frac{\pi}{2}$. The positive definite matrix tries to keep the vector within a certain half space containing the vector. This is analogous to what a positive number does to a real variable. Multiply it and it only stretches or contracts the number but never reflects it about the origin.