Consider the original formulation of the Lasso regression problem in a linear regression setting, as following$$ \min_\beta \|y - X \beta\|_2^2 \ \\s.t. \|\beta\|_1 \leq s $$ To do the optimization, we utilize the Lagrange multiplier, and reformulate the problem as follows, $$ \min_\beta \|y - X \beta\|_2^2 + \lambda \|\beta\|_1 \ $$ From the two formulations, you can see the connection between $\lambda$ and $s$.

(1) as $s$ becomes infinity, the problem becomes unconstrained problem, or ordinary least squares. Thus $\lambda$ becomes 0 accordingly;

(2) as $s$ becomes 0, all $\beta$'s shrink to 0, easily seen from first formulation. Therefore $\lambda$ would go to infinity.

That said, $\lambda$ and $s$ have reverse relationship. Now for your questions.

1. How this constraining parameter $s$ is chosen?

In practice, you would just need to choose $\lambda$, mainly by cross-validation, as other people pointed out. You are not bothered by what the $s$ value would be.

2. How are $\lambda, \, s$, and $\hat{\beta}$ shrinking to zero related to each other?

Have answered by Point (2) I made above.

3. What is the decision process or how are some $\hat{\beta}$'s shrunk to zero and some are not?

This has to do with the L1 constraint.I highly recommend the geometric representation of this problem at P71 of the book The element of statistical learning. The L1 constraint makes the feasible region to be a diamond (in terms of two $\beta$'s, as in the figure). The corners of the region would be "hit" by the function of the residual SS, resulting in shrinking some $\beta$'s to be 0. That's how the sparsity comes from.

Consider the original formulation of the Lasso regression problem in a linear regression setting, as following$$ \min_\beta \|y - X \beta\|_2^2 \ \\s.t. \|\beta\|_1 \leq s $$ To do the optimization, we utilize the Lagrange multiplier, and reformulate the problem as follows, $$ \min_\beta \|y - X \beta\|_2^2 + \lambda \|\beta\|_1 \ $$ From the two formulations, you can see the connection between $\lambda$ and $s$.

(1) as $s$ becomes infinity, the problem becomes unconstrained problem, or ordinary least squares. Thus $\lambda$ becomes 0 accordingly;

(2) as $s$ becomes 0, all $\beta$'s shrink to 0, easily seen from first formulation. Therefore $\lambda$ would go to infinity.

That said, $\lambda$ and $s$ have reverse relationship. Now for your questions.

1. How this constraining parameter $s$ is chosen?

In practice, you would just need to choose $\lambda$, mainly by cross-validation, as other people pointed out. You are not bothered by what the $s$ value would be.

2. How are $\lambda, \, s$, and $\hat{\beta}$ shrinking to zero related to each other?

Have answered by Point (2) I made above.

3. What is the decision process or how are some $\hat{\beta}$'s shrunk to zero and some are not?

This has to do with the L1 constraint. I highly recommend the geometric representation of this problem at P71 of the book The element of statistical learning. The L1 constraint makes the feasible region to be a diamond (in terms of two $\beta$'s, as in the figure). The corners of the region would be "hit" by the function of the residual SS, resulting in shrinking some $\beta$'s to be 0. That's how the sparsity comes from.