Consider the mean squared error $$\text{MSE}(x_0) = E_{T}(f(x_0)-\hat{y}_0)^{2}$$ $$ = E_{T}(\hat{y}_0-E_{T}(\hat{y}_0))^{2}+(f(x_0)-E_{T}(\hat{y}_0)^{2})$$
Is the first term after the equal sign in the second line basically saying the following: Get an estimate and find its deviation from the average estimates over all training samples. The estimate $\hat{y}_0$ will depend on the training set which is why we take the expected value.
Also is $E_{T}(\hat{y}_0)$ constant?
Your intuition seems right, but for clarity, I would rewrite your equations as follows,
\begin{align} \text{MSE}(\hat{y}) &= E\left[(y-\hat{y})^2\right] \\ &= \underbrace{E\left[(\hat{y}-E[\hat{y}])^2\right]}_{\text{Var}[\hat{y}]} + \underbrace{(y-E[\hat{y}])^2}_{\text{Bias}^2} \end{align}
Here $y$ is the actual value of whatever we are estimating, and $\hat{y}$ is its estimator (the value of $\hat{y}$ will depend on the data).
The first term is the variance of the estimator (which is more or less what you said). Also, $E[\hat{y}]$ is non-random (this may be what you meant by constant).
