It's helpful if you cite the paper in which you saw something that you're asking a question about- we could provide a better answer if we knew where the question came from.

First, assume without loss of generality that $A$ and $C$ are symmetric matrices. It's easy to take these quadratic forms and write them in terms of symmetric matrices.

I believe that this problem transformation requires that $A$ be a positive semidefinite matrix- see below.

We'll begin with your second problem and show that it is equivalent to the original problem. We begin with

$\min \mbox{tr}(AB) $

subject to

$\mbox{tr}(CB)=1$

$ \beta \geq 0 $

$\left[
\begin{array}{cc}
1 & \beta^{T} \\\
\beta & B
\end{array}
\right]
\succeq 0
$

By Schur's theorem, the constraint

$\left[
\begin{array}{cc}
1 & \beta^{T} \\\
\beta & B
\end{array}
\right]
\succeq 0
$

is equivalent to

$B \succeq \beta \beta^{T} $

Note that we've implicitly restricted $B$ to being a symmetric matrix.

Next, write $B$ as

$B= \beta \beta^{T} + LL^{T} $

where $L$ is the (slightly generalized) Cholesky factor of $B- \beta \beta^{T}$. If $B-\beta \beta^{T}$ is singular, then $L$ would be singular or even $0$.

Then our problem is equivalent to

$\min \mbox{tr}( A ( \beta \beta^{T} + LL^{T} ) ) $

subject to

$\mbox{tr}(CB)=1$

$ \beta \geq 0 $

$ B=\beta \beta^{T} + LL^{T} $

By the property $\mbox{tr}(DEF)=\mbox{tr}(FDE)$, this problem is equivalent to

$\min \mbox{tr}(\beta^{T}A\beta + L^{T}AL) $

subject to

$\mbox{tr}(CB)=1$

$ \beta \geq 0 $

$ B=\beta \beta^{T} + LL^{T} $

Note that because $A \succeq 0$, $\mbox{tr}(L^{T}AL) \geq 0$ for all $L$, and the term is minimized when $L=0$.

Thus our problem is equivalent to:

$\min \mbox{tr}(\beta^{T}A\beta) $

subject to

$\mbox{tr}(CB)=1$

$ \beta \geq 0 $

$ B=\beta \beta^{T} $

Since $B=\beta \beta^{T}$, and by the cyclic property of $\mbox{tr}()$, this is equivalent to

$\min \mbox{tr}(\beta^{T}A\beta) $

subject to

$\beta^{T}C\beta=1$

$ \beta \geq 0 $

This was your original problem.