(This answer is basically the same as Dan Doel's. We wrote them in parallel, but he posted first. I decided this may still be helpful.)
Unsurprisingly, no one's wrong.
In the nLab article for hyperdoctrines, it talks about "contexts" and classical logic being the case where $T$ is a "category of contexts". For something like first-order logic, what we call the "context" has two parts, usually left implicit. There's a variable context that tells us which free variables are allowed and what their sorts are, and then there are the assumptions. I'm going to focus on the variable contexts, though in some cases these things mix together. As an example, imagine a multi-sorted theory of vector spaces. If we want to say $r(\mathbf v+\mathbf u)=r\mathbf v+r\mathbf u$ we need to specify that $r$ is a scalar, and $\mathbf v$ and $\mathbf u$ are vectors. We could write this like: $r:S,\mathbf v:V,\mathbf u:V\vdash r(\mathbf v+\mathbf u)=r\mathbf v+r\mathbf u$, where $S$ and $V$ are the two sorts of our theory.
The question now is how do we represent variable contexts. The most obvious choice is to think of them like tuples of sorts. We can drop the variable names and just use their positions, giving us something like an object like $S\times V\times V$ identifying the variable context for the above formula. If we want to "forget" the scalar, we can use the projection $S\times V\times V\to V\times V$. When we use a functor to map this category, which we can define to be the category with finite products freely generated by a set of objects (which represent the sorts), we find that it should be contravariant. It's not clear how to turn a formula like $r(\mathbf v+\mathbf u)=r\mathbf v+r\mathbf u$ into one that doesn't have $r$ free, but it's clear how to turn a formula like $\mathbf u+\mathbf v=\mathbf v+\mathbf u$ into one that does have $r$ free. As another example, given the diagonal $V\to V\times V$, we can turn a formula with two variables free, e.g. $\mathbf u-\mathbf v = 0$, into a formula with one variable free, i.e. $\mathbf u-\mathbf u=0$.
Now, in the single-sorted case, we can simplify notation by writing $S\times S$ as $S^2$ and so forth. The diagonal arrow $S\to S^2$ can then be described via $S^{!_2}$ which is the pre-composition of $S$ with $!_2 : 2\to 1$ in the category $\mathbf{FinOrd}$. ($\mathbf{FinOrd}$ is the skeleton of $\mathbf{FinSet}$ and thus they are equivalent.) In other words, we have a contravariant functor $\mathbf{FinOrd}$ to $T$. In fact, $\mathbf{FinOrd}$ is the category with finite coproducts freely generated by a single object. In other words, if we define $T$ to be, in the single-sorted case, the category with finite products freely generated by a single object, then $T\simeq\mathbf{FinOrd}^{op}$. This is basically what Baez is doing. He chooses $T=\mathbf{FinSet}^{op}$ producing $T^{op}\to\mathbf C$ or $\mathbf{FinSet}\to\mathbf C$.