There is an important distinction to make. The second-order induction axiom is just that - an axiom. It can be interpreted in several ways.
In normal second-order arithmetic, $Z_2$, we do have the usual second-order induction axiom
$$
(\forall X)\big [(0 \in X \land (\forall n)[n \in X \to n+1\in X]) \to (\forall n)[n \in X]\big ]
$$
But $Z_2$ is usually studied with first-order semantics, and in that context it is an effective theory of arithmetic subject to the incompleteness theorems. In particular, $Z_2$ includes every axiom of PA, and it does include the second-order induction axiom, and it is still incomplete.
Therefore, the well-known categoricity proof must not rely solely on the second-order induction axiom. It also relies on a change to an entirely different semantics, apart from the choice of axioms. It is only in the context of these special "full" semantics that PA with the second-order induction axiom becomes categorical.
Now, if we fix a sound deductive system for $Z_2$, the change of semantics has no effect whatsoever on the formulas that are provable. So, even though $Z_2$ with full second-order semantics is categorical, for any sound effective deductive system there are still true formulas of $Z_2$ that are neither provable nor disprovable in that system.
This answers a question in the comments, "How can a categorical theory be incomplete?" The answer is that categoricity is determined both by the choice of axioms and the choice of semantics, while completeness in this sense is determined by the axioms and the choice of a deductive system. (Here "complete" means that the set of provable theorems is a maximal consistent set.) There is no reason why, in a general setting, categoricity should imply completeness. In fact, it doesn't.
No matter what semantics we wish to use, it is simply impossible - by the incompleteness theorems - to come up with any effective deductive system that extends PA and is complete in the sense of the previous paragraph. Moving to higher-order systems helps us prove additional true propositions about the natural numbers, but we can never find a deductive system to prove all of them.
If $M\models$ PA, the induction scheme of PA implies that every definable (nonempty) subset of $M$ has a minimal element:
Suppose $\varphi$ is a formula in the language of arithmetic; we want to show that either $\varphi^M$ is empty or $\varphi^M$ has a minimal element.
So let's suppose $\varphi^M$ has no minimal element. Let $\psi(x)\equiv \forall y\le x(\neg\varphi(y))$. Then:
$M\models\psi(0)$, since otherwise $0$ would be the minimal element of $\varphi^M$.
If $M\models\psi(n)$, then $M\models\psi(n+1)$: the only way we could have $\psi(n+1)$ fail in $M$ if $\psi(n)$ holds in $M$ would be for $\varphi(n+1)$ to hold in $M$, in which case - since $\psi(n)$ holds in $M$ - $n+1$ would be the least element of $\varphi^M$.
But now we can apply the induction scheme of PA - with formula $\psi$ - to conclude that $M\models\forall x\psi(x)$. And this means that $\varphi^M$ is empty.
OK, technically in the above I've only talked about parameter-freely definable sets. But it's easy to fold parameters into the argument above.
This means that if $M\models$ PA is nonstandard, then while in reality there are subsets of $M$ which have no minimal element, no such subset of $M$ is definable in $M$. That is: a nonstandard model of PA is "internally" well-founded, but "externally" ill-founded.
To get a sense of how the external-versus-internal distinction behaves, it might be easier to first consider a toy example. Let $\Sigma$ be the language consisting of a single unary function symbol $succ$ (which we'll think of as "successor"), a single binary relation symbol $<$ (self-explanatory), and a single constant symbol $0$ (self-explanatory). Now consider the $\Sigma$-theory $T$ consisting of:
"$succ$ is successor:" $\forall x(x<succ(x))\wedge\forall x,y(\neg(x<y\wedge y<succ(x))) .$
The induction scheme: for each formula $\varphi(x)$ in the language $\Sigma$, we have the axiom $$\varphi(0)\wedge\forall x[\varphi(x)\implies\varphi(succ(x))]\implies\forall x(\varphi(x)).$$ It's easy to show that there is a model $M$ of $T$ which "looks like $\mathbb{N}+\mathbb{Z}$" - concretely, one example of such a model is the following:
The domain of $M$ is all the integers except the negative even integers.
$<^M$ is given by: $$0<2<4<6<...\quad ...<-5<-3<-1<1<3<5<...,$$ and $succ^M$ is the successor operation with respect tot his ordering. The even nonnegative integers form the "$\mathbb{N}$-part" of $M$, and the odd integers form the "$\mathbb{Z}$-part" of $M$.
It is not trivial, but not hard either, to show $M\models T$. As a consequence, analogously to the argument above for PA every nonempty definable subset of $M$ has a minimal element.
However, clearly $M$ has external subsets with no minimal element - e.g. the "$\mathbb{Z}$-part." And the key point is that such external sets need not be hard to describe. There's nothing "absolutely" mysterious about these non-internally-definable sets without minimal elements; they're just mysterious from the point of view of the model itself.
Best Answer
Ind(A(z)) is literally just the statement of mathematical induction, in the special case of the predicate A.
The two proofs you listed are the same proof, except that the chains of equalities are depicted in the reverse order.