Finding zeros of maps between manifolds with different dimensions One of the questions for which the notion of degree is useful is: does this map have a zero. 
For example, one can prove the Fundamental Theorem of Algebra using the following fact involving the degree (c.f. page 110 in Guillemin and Pollack):

Proposition. Suppose that $f:X\rightarrow Y$ is a smooth map of compact oriented manifolds having the same dimension and that $X=\partial W$ ($W$ compact). If $f$ can be extended to all of $W$, then $\text{deg}(f)=0$.

Another standard example would be something like: 

(Exercise II.4.1 in G&P) Show that the function
  $$f(z)=z^7+\cos(|z|^2)(1+93z^4)=0 $$
  has a zero.

A typical solution to this variety of problem involves supposing that $f$ has no zeros so the map $f/|f|$ makes sense everywhere, restricting it to a circle, showing it is homotopic to $z^7$, concluding it has nonzero degree, and then noticing that this contradicts the fact that the map has an extension to the ball bounded by the circle (the Extension Theorem, page 145 in G&P, says it can't be extended if the degree is nonzero). Therefore, $f$ has zeros.
What I am wondering is: is there something that allows one to do something similar in the case when degree theory isn't applicable? For example, if one has a map $F:S^n\rightarrow\mathbb{R}^m$ for $n\neq m-1$ (to avoid examples reducing it to a map $S^n\rightarrow S^n$, where the degree would be defined) can you determine if it has zeros?
 A: The application of degree theory to maps from the plane to itself, such as the exercise from G&P that you quote, come down to the topological fact that $\pi_1(S^1)$, the first homotopy group of $S^1$, is isomorphic to the infinite cyclic group $\mathbb{Z}$---this is the basis of the concept of "winding number".
If I may address for a moment the case of maps of the form $F : S^n \to \mathbb{R}^{n+1}$ (which you have disallowed in your question), the applications of degree theory come down to the fact that $\pi_n(S^n) \approx \mathbb{Z}$; here $\pi_n$ is the $n^{th}$ homotopy group.
For $m,n$ in the range $n < m-1$, every continuous map $F : S^n \to \mathbb{R}^m - 0$ is homotopic to a constant, because $\pi_n(S_{m-1})$ is the trivial group in this range. So in this range the answer to your question "is there something that allows one to do something similar…" is "no". To be specific, there is no topological method of this kind which will allow you to prove existence of a zero of a continuous function of the form $R^{n+1} \to R^m$. This is intuitively clear because since $n+1<m$, any given function which does have a zero at a certain point $p \in R^{n+1}$ can (assuming the zero at $p$ is isolated) be perturbed to remove the zero at $p$.
However, if $n > m-1$ then things get interesting. For instance, $\pi_3(S_2) = \mathbb{Z}$. This is the basis of the Hopf invariant of a 2-component link in $S^3$: given two disjoint, embedded, oriented circles, their Hopf invariant is an integer which is defined by writing down a certain continuous function $S_3 \to S_2$ and taking the corresponding element of $\pi_3(S_2)$. For instance, two ordinary round circles which are "linked" in the ordinary intuitive sense are linked in the rigorous sense that there is no isotopy which pulls those two circles apart; the proof is to compute the Hopf invariant of those two circles and verify that it is $\pm 1$ (depending on orientation).
