Your question is:

Let $S \subset \mathbb{P}^N$ be the image of the Segre map
$\mathbb{P}^n \times \mathbb{P}^n \to \mathbb{P}^N$.
Let $Z \in \operatorname{Sec}_k(S)$. Does $Z$ have rank at most $k$?

Yes. This is *matrix rank*. The elements of $\mathbb{P}^N$ are $(n+1) \times (n+1)$ matrices (up to a scalar factor) and rank with respect to $S$ is ordinary matrix rank. Every element of $\operatorname{Sec}_k(S)$ has rank at most $k$.

A modified question is:

Let $S \subset \mathbb{P}^N$ be the image of the Segre map
$\mathbb{P}^{n_1} \times \dotsm \times \mathbb{P}^{n_s} \to \mathbb{P}^N$.
Let $Z \in \operatorname{Sec}_k(S)$. Does $Z$ have rank at most $k$?

When $s \geq 3$, no. A tensor of border rank $k$ may have rank strictly greater than $k$ or $k+1$. An example is below.

In the symmetric case, the answer is yes for Veronese varieties of degree $2$ (corresponding to quadratic forms), and no for Veronese varieties of degree $d \geq 3$. For example, the homogeneous binary form $x y^{d-1}$ has rank $d$, but lies on $\operatorname{Sec}_2(V)$:
$$
x y ^{d-1} = \lim_{t \to 0} \frac{(tx+y)^d - y^d}{dt},
$$
where each $(1/dt)((tx+y)^d - y^d) \in \operatorname{Sec}_2(V)$.

In the antisymmetric case I believe the answer is again no.

Landsberg's book is a good reference. See also more recently https://arxiv.org/abs/1811.12725 and https://arxiv.org/abs/1812.10267.

For an example of a tensor with border rank $k=2$ and rank greater than $2$ take
$$
T = x y^{d-1} = x \otimes y \otimes \dotsm \otimes y + y \otimes x \otimes y \otimes \dotsm \otimes y + \dotsb + y \otimes \dotsm \otimes y \otimes x,
$$
where $x,y \in \Bbbk^2$ are a basis. It visibly has rank at most $d$ and you can prove that it has rank equal to $d$. It has border rank $2$, since $T = \lim_{t \to 0} ((tx+y)^{\otimes d} - y^{\otimes d})/t$.