Definition: A subbundle $W \subseteq V$ of vector bundle $V \rightarrow X$ is a union vector subspaces $\bigsqcup_x W_x$ which is locally trivial with the subspace topology.
Lemma: A subbdunel $W \subseteq V$ has adapted charts, i.e. for each $x \in X$, there is a neighborhood $U$ and a bundle chart $V|_{U} \cong U \times \Bbb K^n$ that sends $W|_U$ to $U \times \Bbb K^m$
Claim: Let $V \subseteq X \times \Bbb K^n$ be a subbundle, and let $P_x$ be the orthogonal projection onto $V_x$ then $X \mapsto Mat_{n,n}(\Bbb K)$, $ x\mapsto P_x$ is continuous.
The proof of this claim goes as follows:
Proof: The problem is local, so assume $V \cong X \times \Bbb K^m$ using lemma. This isomoprhism defines sections $s_1, \ldots, s_m$ of $V$ which are everywhere linearly independent. By Gram-Schimdt, $s_1, \ldots, s_m$ defines orthonormal basis $t_1, \ldots, t_m$ of $V$.
It is from now on that I do not understand at all:
The inclusion $V \rightarrow X \times \Bbb K^m$ is given by continuous function $A:X \rightarrow Mat_{n,m}(\Bbb K)$ that takes values in the matrices $A$ such that $A^*A=1_m$. The orthogonal projection is $P=AA^*$.
I believe I understand why $A^*A=1_m$ . As a matrix $A$ has rows the $m$ orthonormal vectors. Hence $A^*A$ is precisely the dot product giving $1_m$. But why is $AA^*$ the orthogonal projection?
EDIT: I believe I have obtain the answer.
Let us suppose that the orthonormal basis is given $$\alpha_i = \sum_{k=1}^n \alpha_{ik} e_k$$ for $i=1,\ldots, m$. So to project $e_1$ on to this space we take the inner product,
$$ \sum \langle e_i, \alpha_{i}\rangle \alpha_i$$
So the matrix $A$ is given by $(\alpha_{ki})$. What $AA^* e_1$ does is it gives $$A(\alpha_{11} \, \alpha_{21} \, \cdots \, \alpha_{m1} )^T = \sum_{j=1}^m \alpha_{1j} \alpha_j $$
which is what we wanted.
Best Answer
Since the question you're asking isn't really about the bundle stuff, I'll just ignore that part. Also for simplicity of notation, I'll use the field $\Bbb{R}$, but everything works identically over the complex numbers.
Simplifying, the part you're asking about says, if we have $V\cong \newcommand\RR{\Bbb{R}}\RR^m$ and a metric preserving inclusion $\iota : \RR^m\to\RR^n$, which has matrix $A$ with respect to the standard bases of $\RR^m$ and $\RR^n$, why is $AA^*$ the projection?
Well, what is $A^*$? It's the map $\RR^n\to \RR^m$ such that $\langle A^*v,w\rangle = \langle v, Aw\rangle$, and if $w=e_i$, then $Aw=t_i$ by definition of the map $A$, so $\langle A^*v,e_i\rangle = \langle v, t_i\rangle$. Hence $$A^*v = \sum_i \langle v,t_i\rangle e_i,$$ and $$AA^*v = \sum_i \langle v,t_i\rangle t_i,$$ which is precisely orthogonal projection onto the subspace spanned by the $t_i$, i.e. $V$.
Edit And I see you've figured it out.