An intuitive justification - Metric Embedding Based Approximation Algorithms
The way I remember it is that you also get a lower bound on the expected cost in the embedding. Namely for any two points $x$, $y$ at distance $d(x,y)$ in the original metric, their distance $d&#39;(x,y)$ in the embedding satisfies $d(x,y) \leq d&#39;(x,y)$. If I'm not mistaken, the first lecture mentions that too.

Then, everything is straightforward.

Look at the optimal TSP instance $I$ of the graph $G$. Then consider the embedding of $I$ (call it $I'$) into the tree metric. $I&#39;$ is also an instance of TSP for the tree metric, and its cost is bounded by $E[cost(I&#39;)] \leq \alpha \cdot cost(I)$. 

Therefore, if you compute $J&#39;$, which is the optimal TSP for the tree metric, $E[cost(J&#39;)] \leq E[cost(I&#39;)] \leq \alpha \cdot cost(I)$.

Lift $J&#39;$ back into your initial graph, which gives you an instance $J$. Thus, $cost(J) \leq E[cost(J&#39;)]$. And thus, using the previous chain of inequalities,  $cost(J) \leq \alpha \cdot OPT$.