I have the equation $RSS\left ( \beta \right )=\left ( y-X\beta \right )^{T}\left ( y-X\beta \right )$ that I'm trying to differentiate w.r.t. $\beta$. I used chain rule to get to $\frac{\partial RSS(\beta) }{\partial \beta } = (-X^{T})(y-X\beta)+(y-X\beta)^{T}(-X)$, but I don't know how to get from this to $X^{T}(y-X\beta)$. How do I deal with all the different transposes?

You can proceed as follows:

\begin{align} RSS\left ( \beta \right )&=\left ( y-X\beta \right )^{T}\left ( y-X\beta \right )\\ &=y^Ty - 2y^TX\beta + \beta^TX^TX\beta. \end{align}

Now, if you differentiate $RSS\left ( \beta \right )$ with respect to $\beta$, the derivative will be: $-2 X^T y + 2X^T X\beta$.

The results, which also can be found in any standard books, that I have used are:

1. $\frac{d}{dx}x^TAx = 2Ax$, if $A$ is symmetric.

2. $\frac{d}{dx}y^Tx = y$.

Here $x$ and $y$ denote two different vectors of the same dimension and $A$ is a square matrix of conformable dimension. Furthermore, $x^T$ indicates the transpose of the vector $x$.