If $S \subset X$, with $\partial S = S \cap \partial X$. Must $T(S \cap \partial X) = T(S) \cap T(\partial X)$?

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Some background. This came about in the proof about boundary orientation $$\partial f^{-1}Z = (-1)^{\operatorname{codim}Z} (\partial f)^{-1}Z.$$

A reference would be Guillemin-Pollack page 101.

Denote $S = f^{-1}(Z) \subset X$ where $f: X \to Y$ with $f \pitchfork Z$ and $\partial f \pitchfork Z$. It is remarked that the (orthogonal) complementary space $H$ to $T(\partial S)$ is disjoint from $T(S)$ because $T(S) \cap T(\partial X) = T(\partial S)$.

Basically, why do we have $T(S) \cap T(\partial X) = T(\partial S)$ in the first place? This appears only true if $S\pitchfork \partial X$, but is this immediate from $\partial f \pitchfork Z$?