How can primitive recursion be a special case of minimization?
Your suspicion is correct. Without primitive recursion we get a very small class of functions, namely maps of the form $f(\vec{x}) = x_i + m$, possibly with some undefined values.
Let $\mathcal{F}_k$ be the least class of partial maps $f : \mathbb{N}^k \to \mathbb{N}$ closed under $0$, successor, projections, composition and minimization.

Theorem: For every $f \in \mathcal{F}_k$ there exist $i$ and $m$ such that, for all $\vec{x} = (x_1, \ldots, x_k) \in \mathbb{N}^k$, either $f(\vec{x})$ is undefined or $f(\vec{x}) = x_i + m$.

Proof. We proceed by induction on the structure of $f$. It is obvious that $0$, successor, projections, and compositions preserve the stated property. Now consider $f \in \mathcal{F}_k$ which is defined by minimization of $g \in \mathcal{F}_{k+1}$, so that
$$f(\vec{y}) = \min \{x \in \mathbb{N} \mid g(x, \vec{y}) = 0\}.$$
By induction hypothesis there are two possibilities:

There exists $m$ such that $g(x, \vec{y}) = x + m$ whenever $g(x, \vec{y})$ is defined. In this case:

If $m = 0$ then $f(\vec{y}) = 0$
If $m &gt; 0$ then $f$ is everywhere undefined


There exists $m$ and $i$ such that $g(x, \vec{y}) = y_i + m$ whenever $g(x, \vec{y})$ is defined. In this case:

if $m = 0$ then $$f(\vec{y}) = \begin{cases}0 &amp; \text{if $y_i = 0$ and $g(0, \vec{y})$ is defined}\\ \text{undefined} &amp; \text{otherwise} \end{cases}$$
if $m &gt; 0$ then $f(\vec{y})$ is everywhere undefined.



In all cases minimization preserves the stated property. $\Box$
Let us also try to speculate how the falacy &quot;primitive recursion is a special case of minimization&quot; arises. Suppose $f : \mathbb{N} \to \mathbb{N}$ is defined by primitive recursion by
\begin{align*}
f(0) &amp;= k\\
f(n+1) &amp;= h(n, f(n)),
\end{align*}
where $h : \mathbb{N}^2 \to \mathbb{N}$ is primitive recursive. Then there is a primitive recursive function $g : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ such that
$$
g(x, y) = \begin{cases}
0 &amp; \text{if $f(x) = y$},\\
1 &amp; \text{otherwise.}
\end{cases}
$$
The map $g$ involves a good deal of arithmetic, encoding of lists as numbers, etc. Cruicially, $g$ may be more complicated than $f$ and $h$.
Therefore, we could define $f$ by minimization as
$$f(x) = \min \{y \mid g(x,y) = 0\}.$$
However, we need the entire primitive recursion machinery to get the suitable $g$, so we cannot infer from this that primitive recursion can be dispensed with in the presence of minimization.