Singular value decomposition of a compact operator, using matrix notation Maybe it is a bit naive, but I ask myself the following.
Let $T \in K(X,Y)$ be a compact operator. For example, its singular value decomposition is written this way:
$$Kx=\sum_{n\in \mathbb{N}}{\sigma_n\langle x ,u_n\rangle v_n}$$
But is it forbidden to write it like as if I had matrices of infinite dimensions? Like that:
$$Kx = U\Sigma V^*x$$
Would that break something? Some calculations would come a bit easier and it could avoid with fighting indices.
 A: I presume you intend $X$, $Y$ to be Hilbert spaces.  I'm going to assume we just have one Hilbert space $X = Y$ for convenience.  This doesn't make much difference.
Also I think you probably meant to write $Kx=\sum_{n\in \mathbb{N}}{\sigma_n\langle x ,u_n\rangle v_n}$.
Infinite matrices:  I'm not sure if you know how one associates infinite matrices with bounded linear operators (given a choice of orthonormal basis).  I'm not able to discuss this at length, aside from saying that it does work; moreover, composition of operators corresponds to matrix multiplication,  Hilbert space adjoint of operators to conjugate transpose of matrices, etc.
If you accept this, I can explain how the SVD arises as an operator factorization, which can be translated into a matrix factorization.  Well, sort of explain. You need some background.
You can start with the polar decomposition of $K$,  $$K = U \sqrt{K^* K},$$ where $U$ is a partial isometetry from the closure of the range of $\sqrt{K^* K}$ to the closure of the range of $K$.  This exists for all bounded operators on Hilbert space, no special assumptions needed.  
Since $\sqrt{K^* K}$ is a non-negative self-adjoint compact operator on $X$,  you use the spectral theory of compact operators to obtain an orthnormal basis $u_i \in X$ such that $\sqrt{K^* K} x = \sum_i \sigma_i \langle x, u_i\rangle u_i$, with $\sigma_i$ a decreasing sequence tending to zero.   So put $v_i = U u_i$, and you have the singular value decomposition as you first wrote it. 
So far we don't have any matrices. We have an operator factorization of $K$, $K = U \sqrt{K^* K}$, which, together with the spectral decomposition of $\sqrt{K^* K}$ contains exactly the same information as the SVD, written in your first form, with $u_i$, $\sigma_i$ and $v_i$.
The SVD in your matrix form results from writing matrices corresponding to the operator factorization.
The matrix of   $\sqrt{K^* K}$ with respect to the orthonormal basis $(u_i)$ is the diagonal matrix  $\Sigma$ with diagonal entries $\sigma_i$.
If you have some other favorite orthonormal basis of $X$,  say $(e_i)$,   the matrix of $\sqrt{K^* K}$ with respect to $(e_i)$ is going to be $W^* \Sigma W$ with $W$ the unitary change of basis matrix relating the ON bases $(e_i)$ and $(u)_i$.  Abusing notation I'll still write $U$ for the matrix of the partial isometry $U$ with respect to the ON basis  $(e_i)$.  Now the matrix of $K$ with respect $(e_i)$ is going to be
$$
(U W^*) \  \Sigma \ W,
$$
which is the SVD in the matrix form that you wanted.
