In any scenario/students' choice of toppings, you don't have to order more than $n$ pizzas - each student picks one for themselves with one of their favorite topping.
Another observation - you will always have to order at least $\lceil\frac{n}{2}\rceil$ pizzas, because any student can eat no less than half a pizza. If there weren't any constraints on toppings, what you could do is divide students into groups of $2$ (with one group of size $1$ if $n$ is odd), and let each group order a pizza of their choosing. Note that you wouldn't be able to "move" any of the students to another (fully populated) group, because then they would have to divide their pizza into more parts than is allowed ($2$ in this case). Basically, this is the pigeonhole principle in action.
For example, if you had $n=5$ students and every one of them wanted garlic on their pizza, you will have to order at least $3$ pizzas (even though they will all be the same with garlic as a topping), but you never have to order more than $5$.
With added constraints on toppings, the number of pizzas may be larger than $\lceil\frac{n}{2}\rceil$, but as stated earlier, no greater than $n$.
If you had $n=2$ students, but one of them wanted only garlic, and the other one wanted sausage, you have to order $2$ pizzas to satisfy the constraints (even though $\lceil\frac{n}{2}\rceil=1<2$)
Those observations lead to a formulation of the problem that focuses more on groups of students rather than pizzas themselves, since each group will have to order a distinct pizza anyway; the objective is to find a grouping of students that satisfies topping constraints (each group gets a pizza such that every member gets what they want) and minimizes the number of groups (which is equal to number of pizzas).
More formally:
(Edit: added for accuracy) Let $T = \{t_1, t_2, ...\}$ be a set of all possible toppings. (for example - $T=\{mushrooms,onions,garlic,sausage\}$).
Let $S=\{S_1, S_2, ...,S_n\}$ be the family of non-empty sets of students' favorite toppings; $S_i \in T$ (for example $S_1=\{garlic, onion\}$);
Let $k$ be the maximum number of slices on a pizza (in your original problem, $k=2$, but we can generalize here);
Let $X=\{X_1, X_2,...\}$ be the family of groups of students (or a "grouping"). Each set $X_i$ is a subset of $S$ (you can define $X$ alternatively as a partition of $S$).
Constraints:
$S_i \neq \emptyset$ for each $S_i \in S$ (everyone has at least one favorite topping).
$X_i \cap X_j = \emptyset$ for $X_i, X_j \in X$. (groups are disjoint).
$1 \le |X_i| \le k$ for each $X_i \in X$. (groups can't be empty, but also no larger than $k$).
$\bigcup_{X_i \in X} X_i = S$ (no students are left out).
$\bigcap_{S_j \in X_i} S_j \neq \emptyset$ (each group has a common favorite topping).
We want to find a grouping $X$ that satisfies constraints listed above, such that $|X|$ is minimal:
$$(X=?) \implies (|X| \to min)$$
This model takes into account every constraint you listed. At first glance, the problem looks NP-hard with complexity of roughly $O(n!)$ (think of how many possible groupings there are: you can make $n \choose k$ different groups of size exactly $k$, there are at least $\lceil\frac{n}{2}\rceil$ different groups in a grouping, groups can be of different size, and so on...)
Unless there's some property of the problem I'm missing, you have to search the entire space of groupings to determine the best, hence the problem is computationally hard in the general case. Maybe that's not a problem for small values of $n$?
Also, if you absolutely need this problem to be modelled in terms of a classical problem, you can read about the hitting set problem (this is some material I found online that explains it). My first attempt of modelling your problem used the hitting set problem as a basis, but I ran into issues with constraints regarding $k$. What I found, though, is that the hitting set problem is equivalent to your problem if we don't concern ourselves about $k$ (so if a student can eat any amount of pizza to be happy). Since the hitting set problem is in itself NP-hard, I'm inclined to think that your variant (which has more constraints) is similarly complex as well.
EDIT: Why is the simpler version equivalent to the hitting set problem?
We forget about constraints with $k$, that is - a student can now eat any amount of pizza (no matter if it's half of it, a third, or any fraction) to be satisfied. This allows for a simpler model:
$T = \{t_1, t_2, ...\}$ - set of all possible toppings.
$S = \{S_1, S_2, ..., S_n\}$ - family of sets of preferred toppings, $S_i \in T$.
$P = \{p_1, p_2, ...\}$ - set of pizzas ordered. Let each element also $p_i$ describe a topping of that pizza, so for example $p_3=onion$. (note that each type of topping can be ordered only once and still satisfy constraints, because a pizza can be shared by an arbitrary number of students - there's no need to order two pizzas of the same topping).
We only have one constraint here:
$\forall S_i : S_i \cap P \neq \emptyset$ ($P$ intersects each set of student's choices; this guarantees that each student will find a pizza in $P$ that has one of their favorite toppings)
And our objective is minimizing $|P|$:
$$(P=?) \implies (|P| \to min)$$
In short, we are finding the smallest set $P$ that has at least $1$ common element with every set $S_i$, which is the formulation of the hitting set problem. There is no need to worry about dividing students into groups, because as long as each student can find a pizza they like in $P$, the solution is valid.
As I have written earlier, I didn't find a way to extend the simpler model to the variant where a student has to eat at least half a pizza, that's why in the original answer I opted for another approach, independent of the hitting set problem.
