I will try with an informal approach.
- Tangent space and curves on manifolds
Let $p$ be any point of a given manifold $S$ of dimension $n$. With $\gamma:[a,b]\rightarrow S$ we denote any path on $S$ passing through $p$, with $\gamma(0)=p$.
To understand the tangent space $T_pS$ of $S$ at $p$, one needs to consider the tangent vectors
$$\frac{d(\phi\circ\gamma)}{dt}(0) $$
at $t=0$, denoting by $\phi:U\ni p\rightarrow \mathbb R^n$ any local chart around $p$. The derivative w.r.t. $t$ of $\phi\circ\gamma$ is computed using the standard techniques of Analysis. The tangent space at $p$ consists of the equivalence classes of such paths through $p$; two paths are equivalent if their first derivatives at $0$ coincide. One puts a linear space structure on $T_pS$ and checks that all constructions are chart-independent.
- Tangent space on Stiefel manifolds
If we consider a Stiefel manifold $V_{n,p}$, some simplifications occurr. In fact, any point $Y$ on the Stiefel manifold $V_{n,p}$ is an $n\times p$ matrix which satisfies
$$Y^T Y=I. $$
This implies that any path $\gamma:[a,b]\rightarrow V_{n,p}$ is s.t. $t\mapsto \gamma(t)=Y(t)$, i.e. the image of $\gamma$ is a matrix on the Stiefel manifold with elements which are functions of $t$, for all $t\in[a,b]$. In other words, we can use the standard calculus techniques to compute
$$\frac{d}{dt}\gamma $$
elements-wise in $\gamma(t)\in V_{n,p}$.
In order to find the relations defining the tangent space at $Y\in V_{n,p}$ we follow the above scheme. We begin by considering a path $\gamma$ on the Stiefel manifold s.t. $\gamma(t)=Y(t)$ and $\gamma(0)=Y(0):=Y$; the relations $Y(t)^T Y(t)=I$ are equivalent to
$$\gamma^T(t)\gamma(t)=1(t), ~~(*)$$
denoting by $1(t)$ the image of the constant path on $V_{n,p}$, i.e. $1(t)=I$, for all $t$.
We differentiate $(*)$, obtaining
$$\frac{d}{dt}\left(\gamma^T\gamma\right)=0,$$
and
$$\left(\frac{d\gamma}{dt}(0)\right)^T\gamma(0)+ \gamma(0)^T\frac{d\gamma}{dt}(0)=0$$
at $t=0$. Denoting by $\Delta:=\left(\frac{d\gamma}{dt}(0)\right)^T$ and recalling that, by definition, $\gamma(0)=Y$, we arrive at
$$\Delta^TY+ Y^T\Delta=0. $$
In the reference at pag. 307 you can find the above formula. $\Delta$, by definition, is any element of the tangent space at $Y$. I hope it helps.
- On the projections onto the tangent, resp. normal space at $Y$
Let us consider the setting above. The normal space at $Y$ is the space
$$\mathcal N_Y=\{N\in V_{n,p}: tr(\Delta^TN)=0\}, $$
as the trace map $tr$ defines an inner product, as shown at pag. 308. The important step to show is the following: for any $Z\in V_{n,p}$, then
$$A=\pi_N(Z):=Y\operatorname{sym}(Y^TZ)\in \mathcal N_Y.$$
In fact
$$tr(\Delta^TA)=\frac{1}{2}tr(\Delta^T Y(Y^TZ+(Y^TZ)^T))=\frac{1}{2}\left(tr(\Delta^T Z)-tr(\Delta Z^T)\right)=0,$$
for the properties of trace. Similarly, one proves that $\pi_T(Z)\in T_Y V_{n,p}$, i.e.
$$\pi_T(Z)^T Y+Y^T\pi_T(Z)=0. $$
The final step towards the decomposition is to show that the equality
$$Z=\pi_T(Z)+\pi_N(Z), $$
holds. In other words, each $Z$ is determined by the orthogonal splitting in tangential and normal components.
