If you set $u = \frac{1}{|p|}p$, then
$$(u^Tv)u$$
is the orthogonal projection of $v$ onto the subspace (here a line) spanned by $p$.
Note that $(u^Tv)$ is a scalar and $(u^Tv)u$ can be rewritten as
$$(u^Tv)u = (uu^T)v$$
Now,
$$v- (uu^T)v = (I-uu^T)v$$
means subtracting the component of $v$ in direction of $p$ to obtain the remaining component of $v$ in the direction orthogonal to $p$.
(Here you may think, for example, of the decomposition of a force or velocity or any other vector $v$ into two orthogonal components, where the direction $p$ of one component is already given.)
Edit after comment (setting up the matrix):
$$u = \frac{1}{\sqrt{5}} \begin{pmatrix} 1 \\ 2\end{pmatrix} \Rightarrow uu^T = \frac{1}{\sqrt{5}} \begin{pmatrix} 1 \\ 2\end{pmatrix} \cdot \frac{1}{\sqrt{5}} \begin{pmatrix} 1 & 2\end{pmatrix} = \frac{1}{5} \begin{pmatrix} 1 & 2 \\ 2 & 4\end{pmatrix}$$
So, you get
$$I - uu^T = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} - \frac{1}{5} \begin{pmatrix} 1 & 2 \\ 2 & 4\end{pmatrix} = \frac{1}{5}\begin{pmatrix} 4 & -2 \\ -2 & 1\end{pmatrix} $$
You can check, that this matrix $P = I - uu^T$ has the property $P^2 = P$, hence is idempotent.