Are the vertical sections of the Ackermann function primitive recursive?

No, already $A(n,3)$ is not primitive recursive. Let me use the essentially equivalent up-arrow notation: $A(n,m)=2\uparrow^{n-1}m$, and argue why $f(n)=2\uparrow^n 3$ is not PR. I claim $f(2n-2)\geq 2\uparrow^{n+1}n=A(n,n)$. $n\mapsto A(n,n)$ outgrows all $n\mapsto A(m,n)$, so by a classical argument it eventually outgrows all PR functions, and hence so will $f$ after we justify the claim.

It is enough to show that for $m\geq 3$ we have $2\uparrow^n m\geq 2\uparrow^{n-1}(m+1)$. We show this by induction on $m$. Indeed, $2\uparrow^n m=2\uparrow^{n-1}(2\uparrow^n(m-1))$, so it is enough to show $2\uparrow^n(m-1)\geq m+1$. This holds true for $m=3$ as $2\uparrow^n 2=4$ for all $n$, and for larger $m$ we have $2\uparrow^n(m-1)=2\uparrow^{n-1}(2\uparrow^n(m-2))\geq 2\uparrow^{n-1}m\geq m+1$.

Let me note that the bound $f(2n-2)\geq 2\uparrow^{n+1}n$ is far from optimal, in fact it should hold for $f(n+2)$, at least for large enough $n$.

If you don't mind not getting the best possible result, then the proof is very short. Indeed,

$$A(n,4)=A(n-1,A(n,3))$$

so if you can prove that $A(n,3)\geq n-1$ then $A(n,4)\geq A(n-1,n-1)$, which we know doesn't grow primitive-recursively. But $A(n,3)$ grows much more quickly than $n-1$.