Definition of the degree of a morphism $\mathbb{P}^1 \rightarrow \mathbb{P}^r$ First of all, I must thank you for taking the time to read this, since it is actually a question with multiple parts, and it's kind of long.
I am reading Kock and Vainsencher's An Invitation to Quantum Cohomology. In the very first page of Chapter 2 on Stable Maps, the following definition is given: (we work over $\mathbb{C}$)

Definition: By the degree of a map $\mu:\mathbb{P}^1\rightarrow \mathbb{P}^r$ we mean the degree of the direct image cycle $\mu_*[\mathbb{P}^1]$. In particular, a constant map has degree zero. In other words, if $e\geq 1$ is the degree of the image curve (with reduced scheme structure), and $k$ denotes the degree of the field extension corresponding to the map, then the degree of the map is $k\cdot e$. Note that, except for the case in which the image curve is a straight line, the definition above differs from the usual definition, given just by the degree of the field extension.

Then, the following is statement follows immediately:

To give a map $\mu:\mathbb{P}^1\rightarrow \mathbb{P}^r$ of degree $d$ is to specify, up to a constant factor, $r+1$ binary forms of degree $d$, which are not allowed to vanish simultaneously at any point. This condition defines a Zariski open subset
  $$
W(r,d) \subset \mathbb{P}\left(\bigoplus_{i-1}^r H^0(\mathcal{O}_{\mathbb{P}^1}(d))\right)
$$

Now, for a book that claims to only require Chapter 1 of Hartshorne, all the above seem very cryptic. Specifically, my questions are the following:


*

*From what I have read on various other sources, the direct image cycle mentioned comes from the following map: 
$$
\mu_*: Z_k(\mathbb{P}^1) \rightarrow Z_k(\mathbb{P}^r)
$$
of the abelian groups of $k$ cycles, however, I am struggling to find consistent definition on the constructions of these groups and the above map, like do we only consider 1-cycles in our case because $\mathbb{P}^1$ is irreducible? More specifically, how is the direct image cycle actually defined? And what is meant by its degree?

*How does the definition imply what follows "in other words..." about the degree of the field extension corresponding to the map?

*Which field extension is the field extension corresponding to the map? Is it $[\mathbb{C}(\mathbb{P}^r):\mathbb{C}(\mu(\mathbb{P}^1))]$? or $[\mathbb{C}(\mathbb{P}^1):\mathbb{C}(\mathbb{P}^r):]$? (I've seen both in different sources, specifically, the first one from Shafarevich and Danilov's Algebraic Geometry I: Algebraic Curves, Algebraic Manifolds and Schemes, Chapter II on Algebraic varieties and schemes Section 5; the second one from Hartshorne Chater II Section 6, pg. 137)
Do either of them make sense? 

*What exactly is the "usual definition"? And how does it differ from the definition given here?

*How does the definition given connect with the statement following it (the one on binary forms)? In particular what are all the objects in the inclusion statement? What is this  $\mathbb{P}\left(\bigoplus_{i-1}^r H^0(\mathcal{O}_{\mathbb{P}^1}(d))\right)$ ?
Basically, this entire page does not make much sense to me, and I suppose all of the above points would be resolved from a complete explanation of the definition and what comes after it. My understanding of algebraic geometry is roughly at around beginning of Chapter 2 of Hartshorne on sheaves (As I mentioned before the author indicates that knowledge of Chapter 1 is sufficient, and one does not lose anything by considering schemes as varieties).
Any help is appreciated, and thanks for taking the time to read all this!
 A: *

*If you're more comfortable with singular homology, then I think it's helpful to think of singular homology instead. In this case the pushforward map is just the usual pushfoward map in homology $H_*(\mathbb{P}^1,\mathbb{Z}) \rightarrow H_*(\mathbb{P}^r,\mathbb{Z})$. If you want to stick with talking about cycles, then the (proper) pushforward map is just defined by mapping an irreducible subvariety $[V]$ to its image $[f(V)]$. Note that you need the properness assumption here to ensure that $f(V)$ is closed. It is a theorem that this induces a map on Chow groups $A_*$, so in this case we have $A_1(\mathbb{P}^1) \rightarrow A_1(\mathbb{P}^r)$.

*The usual definition of the degree of a map between two varieties of the same dimension $V_1 \rightarrow V_2$ is given by the degree of the corresponding field extension.  

*There is another notion of a degree - this is defined for subvarieties of $\mathbb{P}^r$. Given a subvariety $V\subset \mathbb{P}^r$ of dimension $k$, its degree is defined to be the number of points you get after interesting $V$ with $n-k$ generic hyperplanes. For intuition, think in the case of curves in $\mathbb{P}^2$. Intersecting a degree $d$ curve (a curve cut out by a degree $d$ homogeneous equation) with a line gives $d$ points. 

*The definition of degree here is the product of the above two notions of degree. You can think of this number as the number of points in the pre-image of the set of number of points you get by intersecting your curve with a generic hyperplane. 

*So for example, the closed embedding $\mathbb{P}^1 \hookrightarrow \mathbb{P}^2$ as a conic is a map of degree $2$. The map $\mathbb{P}^1 \rightarrow \mathbb{P}^1$ given by $[X:Y] \mapsto [X^2:Y^2]$ is a map of degree $2$. If you compose these two maps, you get a map $\mathbb{P}^1 \rightarrow \mathbb{P}^1 \hookrightarrow \mathbb{P}^2$ of degree $4$. Your image is a conic. Intersecting the conic with a hyperplane gives 2 points. Each point has 2 points in the pre-image (well, generically) in the first map (the 2:1 cover). 

*This notion of degree, for a map $f:\mathbb{P}^1 \rightarrow \mathbb{P}^r$, also turns out to be also equal to $c$, where $f_*[\mathbb{P}^1] = c[L]$, where $[L] \in H_2(\mathbb{P}^r,\mathbb{Z})$ is the generator (the class of a line). You can replace $H_2$ with $A_1$ if you like. (Note that $H_2(\mathbb{P}^r,\mathbb{Z})$ / $A_1(\mathbb{P}^r)$ is isomorphic to $\mathbb{Z}$). In fact this is probably how I would define it in the first place.

*A map from $\mathbb{P}^1 \rightarrow \mathbb{P}^r$ is always given by
a $r+1$ tuple of binary forms of the same degree. To see that the
degree of such a bilinear form has to be equal to $d$, you can use
for example the equivalent definition above that the pre-image of the
intersection of the image of your curve and a generic hyperplane
should have $d$ points.

