It is relatively simple to derive the Schmidt decomposition of a pure state $|{\psi}\rangle \in H_A \otimes H_B$ with the SVD decomposition theorem. There are plenty of examples (lecture notes, books, videos, etc.) on the subject.
The question I have is clearly stated in the title. Is it possible to derive a Schmidt decomposition for a mixed state? Such a state is expressed as a set of pure states $|{\psi_i}\rangle$ with probability distribution $p_i$, for $1 \leq i \leq t$.
Usually, the Schmidt rank of a mixed state $\rho$ is defined as the minimum of the maximal Schmidt rank of an element in any decomposition of $\rho$ into a mixture of pure states (there are many different decompositions if $\rho$ is not pure), see B. Terhal and P. Horodecki, A Schmidt number for density matrices.
In general, it's very hard to compute it. The separability problem is equivalent to asking if $\rho$ has Schmidt rank 1 or higher. And it's known that this problem is NP-hard.
Update
A second meaning when we talk about Schmidt decomposition of mixed states is the following. We can consider a mixed state $\rho$ as a vector in the space of matrices $\mathcal{L}(H_A) \otimes \mathcal{L}(H_B)$. And thus we can write down a so-called operator-Schmidt decomposition $$ \rho = \sum_i \gamma_i A_i \otimes B_i, $$ where $\gamma_i \ge 0$ and $\{A_i\}$, $\{B_i\}$ are ONBs in $\mathcal{L}(H_A)$, $\mathcal{L}(H_B)$ respectively.
In general, matrices $\{A_i\}$ and $\{B_i\}$ won't be quantum states. Nevertheless, this decomposition can be useful in recognizing the separability of $\rho$ (see N. Johnston, What the Operator-Schmidt Decomposition Tells Us About Entanglement), as was pointed out by John Watrous in the comments.
Note that this decomposition also defines what is called an operator-Schmidt rank of $\rho$. This is a different number than Schmidt rank (or Schmidt number) of $\rho$ defined earlier.
