Why is $\forall x \in \mathbb{N}\ \Phi(x,x)$ unary but $\Phi$ is binary?
A lot of confusion arises because the sloppy notation and imprecise terminology involving free and bound variables. Let me try to clear some of it up.
The title of your question wrongly claims that $\forall x \in \mathbb{N} \,.\, \Phi(x,x)$ is a function, which it is not. It is a truth value. Let us see why this is the case.
Suppose someone defined $f$ by writing down
$$f(x,y) = x^2 + 3 y,$$
where $x$ and $y$ range over $\mathbb{N}$. First of all, it would be more precise to define $f$ like this:
$$f = ((x, y) \mapsto x^2 + 3 y).$$
The right-hand side is read as &quot;the function which maps $(x,y)$ to $x^2 + 3 y$&quot;. The variables $x$ and $y$ are bound, i.e., we get the same function if we rename them, for example $(u, v) \mapsto u^2 + 3 v$ is equal to $f$.
The function $f$ is binary because it takes two arguments.
With $f$ defined as above:

The expression $f(x, y)$ is not a function! It is a number. Indeed, you would have no trouble claiming that $f(x,y) \in \mathbb{N}$, would you? However, this number is parameterized by two free variables $x$ and $y$. Thus, which number it is depends on $x$ and $y$.

The previous remark maybe becomes clearer if we contemplate the fact that $f(3, 4)$ is not a function but a number, and so is $f(u,u)$ where $u \in \mathbb{N}$ is a free variable.

We can fix just one of the arguments to get a unary function, for example $x \mapsto f(x, 7)$ is unary because it is a map $\mathbb{N} \to \mathbb{N}$.

The expression $f(x, 7)$ is not a function! It is a number, parameterized by the free variable $x$.


Next, consider the definition
$$g = (x \mapsto x^2 + 3 y).$$
The function $g$ is unary because it takes one argument. However, $g$ depends on the free variable $y$. (Some people may call this $y$ a &quot;constant&quot;, or call $x$ an &quot;argument&quot; and $y$ a &quot;parameter&quot;, but let us not do that. We will stick strictly to the terminology &quot;$x$ is a bound variable&quot; and &quot;$y$ is a free variable&quot;.) We may rename the bound variable but not the free one:

$(u \mapsto u^2 + 3 y)$ is equal to $g$
$(u \mapsto u^2 + 3 v)$ is not equal to $g$ because $y$ and $v$ are different free variables
$(y \mapsto y^2 + 3 y)$ is not equal to $g$ because the free variable $y$ does not appear in it.

Some people might define $g$ above like this:
$$g_y = (x \mapsto x^2 + 3 y).$$
They want to indicate that $y$ is a free variable in the definition of $g$. In our notation, this would correspond to
$$g = (y \mapsto (x \mapsto x^2 + 3 y))$$
so that now $g$ is a unary function without free variables. When we apply $g$ to $y$ by writing $g_y$, we again get a unary function, this time wit one free variable $y$. By the way, writing $g_y$ instead of $g(y)$ is just cosmetics.
Consider
$$\Phi = ((u, v) \mapsto u &lt; 7 \lor (v + u^2 &gt; 6).$$
Then $\Phi$ is a binary function because it maps from $\mathbb{N} \times \mathbb{N}$ to $\{\bot, \top\}$. It should be clear that:

$\Phi(x,x)$ is not a function. It is a truth value parametrized by the free variable $x$.

$x \mapsto \Phi(x,x)$ is a unary function mapping from $\mathbb{N}$ to truth values $\{\bot, \top\}$.

$\forall x \in \mathbb{N} \,.\, \Phi(x,x)$ is not a function. It is a truth value. Indeed, we can plug in $\Phi$ and get $\forall x \in \mathbb{N} \,.\, x &lt; 7 \lor (x + x^2 &gt; 6)$, which is false (a truth value, not a function).


What I think you really wanted to ask was this: why is $\Phi$ binary but $x \mapsto \Phi(x,x)$ is unary? Is the answer not obvious, once we use correct notation that does not mess up free and bound variables?
Regarding the discussion with your teacher, there you need to correctly indicate which things are supposed to be free variables and which ones arguments of functions. You wrote:
$$\Phi^n(x_1, \ldots, x_n, y) = \Psi_y^n(x_1, \ldots, x_n).$$
I am sure you can see such thigs in textbooks, but they are a bit sloppy. To be very precise, we need information on which things in $\Psi$ are supposed to be free variables and which ones are arguments. I would guess that what is meant is that
$$\Psi_y^n : \underbrace{\mathbb{N} \times \cdots \mathbb{N}}_n \to \mathbb{N},$$
which means that
$$\Psi : \mathbb{N} \to (\mathbb{N} \to \underbrace{\mathbb{N} \times \cdots \mathbb{N}}_n \to \mathbb{N})).$$
We see that $\Psi$ is a unary function, $\Psi_n$ is a unary function, and $\Psi^y_n$ is an $n$-ary function with free variables $n$ and $y$.
If we read your definition as
$$\Phi = (n \mapsto ((x_1, \ldots, x_n, y) \mapsto \Psi_y^n(x_1, \ldots, x_n)))$$
then $\Phi$ is an unary function and $\Phi_n$ is an $(n+1)$-ary function with free variable $n$.