Some confusion may come from considering the product space $\mathbb{R}^\mathbb{Z}$, rectangles, and projections as central to the problem. They are not.
Let's go back to the beginning. Your first sentence says "Let $X_i$ be a sequence of i.i.d. rv.". This means there is a probability space $(\Omega,{\cal F},\mathbb{P})$ and measurable maps $X_i:(\Omega,{\cal F})\to (\mathbb{R},{\cal B}(\mathbb{R}))$ for $i=1,2,3\dots.$ These maps have other important properties, crucial to the proof of Kolmogorov's $0-1$ theorem, but let's ignore those for now.
You have correctly explained that $B_0:=\sigma(X_1,\ldots)$ is the smallest $\sigma$-algebra that makes all the maps $(X_i)_{i\geq 1}$ measurable. Note that $B_0$ is a subset of $\cal F$. Similarly, $\sigma(X_1,\ldots,X_k)$ is the smallest $\sigma$-algebra that makes all the maps $(X_i)_{1\leq i\leq k}$ measurable, and $B_k$ is the smallest $\sigma$-algebra that makes all the maps $(X_i)_{i\geq k+1}$ measurable. All of these are subsets of $\cal F$; in fact they are all subsets of $B_0$.
So, to address some of your questions:
"After all, it wouldn't make any sense to even consider $\cap B_k$ right?" Sure, why not? Each $B_k$ is a subset of $\cal F$, and $\cap B_k$ means their intersection.
"What is the formal definition of $B_k$?" Answer: $B_k$ is the smallest $\sigma$-algebra that makes all the maps $(X_i)_{i\geq k+1}$ measurable. This is the formal definition.
"Is $B_k$ a projection?" No, $B_k$ is a subset of $\cal F$.
"In other words, do I just consider the smallest sigma algebra generated by $X_{k+1},\ldots$ and then slap on the necessary extra stuff on the left?" Answer: $B_k$ is the smallest $\sigma$-algebra that makes all the maps $(X_i)_{i\geq k+1}$ measurable. Nothing needs to be added to it.
It may be that I have completely missed what you are confused about. If so, let me know. But this should get you started.