The main issue here is that in real problems you don't have f(x) or y_test because it is what you try to predict.
The best way to deal with Bias-Variance trade-off is to analyse the model complexity. For example, in Random Forest you can increase the n_estimators or max_depth increasing the variance and reducing the bias.
As you can see in the image, after max_depth=15 you don't have an improvement in the score. Give a higher value to max_depth will increase the complexity and the variance without a better score, hence 15 could be a good "Trade-Off".
In the other hand, you can split your data set in Train and Test datasets, in this case, you still don't have the f(x) but at least you will have y_test (real values)
It helps to think carefully about exactly what type of objects $\hat \theta$ and $\hat g$ are.
In the top case, $\hat \theta$ would be what I would call an estimator of a parameter. Let's break it down. There is some true value we would like to gain knowledge about $\theta$, it is a number. To estimate the value of this parameter we use $\hat \theta$, which consumes a sample of data, and produces a number which we take to be an estimate of $\theta$. Said differently, $\hat \theta$ is a function which consumes a set of training data, and produces a number
$$ \hat \theta: \mathcal{T} \rightarrow \mathbb{R} $$
Often, when only one set of training data is around, people use the symbol $\hat \theta$ to mean the numeric estimate instead of the estimator, but in the grand scheme of things, this is a relatively benign abuse of notation.
OK, on to the second thing, what is $\hat g$? In this case, we are doing much the same, but this time we are estimating a function instead of a number. Now we consume a training dataset, and are returned a function from datapoints to real numbers
$$ \hat g: \mathcal{T} \rightarrow (\mathcal{X} \rightarrow \mathbb{R}) $$
This is a little mind bending the first time you think about it, but it's worth digesting.
Now, if we think of our samples as being distributed in some way, then $\hat \theta$ becomes a random variable, and we can take its expectation and variance and whatever we want, with no problem. But what is the variance of a function valued random variable? It's not really obvious.
The way out is to think like a computer programmer, what can functions do? They can be evaluated. This is where your $x_i$ comes in.
In this setup, $x_i$ is just a solitary fixed datapoint. The second equation is saying as long as you have a datapoint $x_i$ fixed, you can think of $\hat g$ as an estimator that returns a function, which you immediately evaluate to get a number. Now we're back in the situation where we consume datasets and get a number in return, so all our statistics of number values random variables comes to bear.
I've discussed this in a slightly different way in this answer.
Is it correct to think of this as each observation/fitted value having its own variance and bias?
Yup.
You can see this in confidence intervals around scatterplot smoothers, they tend to be wider near the boundaries of the data, as there the predicted value is more influenced by the neighborly training points. There are some examples in this tutorial on smoothing splines.
Best Answer
When looking at the squared error loss decomposition $$\mathbb{E}_\theta[(\theta-\delta(X_{1:n}))^2]=(\theta-\mathbb{E}_\theta[\delta(X_{1:n})])^2+\mathbb{E}_\theta[(\mathbb{E}_\theta[\delta(X_{1:n})]-\delta(X_{1:n}))^2]$$ I only see two terms: one for the bias and another one for the variance of the estimator or predictor, $\delta(X_{1:n})$. There is no additional noise term in the expected loss. As should be since the variability is the variability of $\delta(X_{1:n})$, not of the sample itself.
My interpretation of the squared bias+variance decomposition [and the way I teach it] is that this is the statistical equivalent of Pythagore's Theorem, namely that the squared distance between an estimator and a point within a certain set is the sum of the squared distance between an estimator and the set, plus the squared distance between the orthogonal projection on the set and the point within the set. Any loss based on a distance with a notion of orthogonal projection, i.e., an inner product, i.e., essentially Hilbert spaces, satisfies this decomposition.
The question is unclear: if by minimum over models, you mean $$\min_\theta \mathbb{E}_\theta[(\theta-\delta(X_{1:n}))^2]$$ then there are many examples of statistical models and associated decisions with a constant expected loss (or risk). Take for instance the MLE of a Normal mean.
In a generic sense, the bias is the distance between the true model and the closest model within the assumed family of distributions. If the true model is unknown, the bias can be ascertained by bootstrap.
When considering another loss function like $$(\theta-\mathbb{E}_\theta[\delta(X_{1:n})])^2+\alpha[(\mathbb{E}_\theta[\delta(X_{1:n})]-\delta(X_{1:n}))^2]\qquad 0<\alpha$$ pushing $\alpha$ to zero puts most of the evaluation on the bias while pushing $\alpha$ to infinity switches the focus on the variance.