For a general kernel it is difficult to interpret the SVM weights, however for the linear SVM there actually is a useful interpretation:

1) Recall that in linear SVM, the result is a hyperplane that separates the classes as best as possible. The weights represent this hyperplane, by giving you the coordinates of a vector which is orthogonal to the hyperplane - these are the coefficients given by svm.coef_. Let's call this vector w.

2) What can we do with this vector? It's direction gives us the predicted class, so if you take the dot product of any point with the vector, you can tell on which side it is: if the dot product is positive, it belongs to the positive class, if it is negative it belongs to the negative class.

3) Finally, you can even learn something about the importance of each feature. This is my own interpretation so convince yourself first. Let's say the svm would find only one feature useful for separating the data, then the hyperplane would be orthogonal to that axis. So, you could say that the absolute size of the coefficient relative to the other ones gives an indication of how important the feature was for the separation. For example if only the first coordinate is used for separation, w will be of the form (x,0) where x is some non zero number and then |x|>0.

Yes, it is the equation of a general hyperplane.

As to the second question: suppose for some $r>0$ we have two hyperplanes $w_r\cdot x_1 + b_r = r$ and $w_r\cdot x_{-1} + b_r = -r$. Take two points, $x_1$ and $x_{-1}$, one on each of the hyperplanes, then we can use the Cauchy-Schwartz inequality to find a lower bound on the distance between these two points. (Note that the distance between $x_1$ and $x_{-1}$ is just $\|x_1-x_{-1}\|$) \begin{align} \frac{2r}{\|w_r\|} &= \frac{(r-b_r) -(-r-b_r)}{\|w_r\|} = \frac{w_r\cdot x_1 - w_r\cdot x_{-1}}{\|w_r\|} =\frac{w_r}{\|w_r\|}\cdot (x_1-x_{-1})\\ &\overset{\text{C.S.}}{\leq} \underbrace{\left\|\frac{w_r}{\|w_r\|} \right\|}_{=\,1}\;\|x_1-x_{-1}\| = \|x_1-x_{-1}\| \end{align}

The margin is the minimum distance from a point on one plane to another. This minimum is obtained (by Cauchy-Schwartz) if $x_1-x_{-1}$ is parallel to $w$ (and therefore perpendicular to the hyperplane, which is geometrically intuitive).

So the margin would have a width of $\frac{2r}{\|w_r\|}$.

However, this form of the SVM may be expressed as $$\text{Minimize}\quad \|w_r\|\quad\text{s.t.}\quad y_i(w_r\cdot x_i+b_r) \geq r\; \text{for $i=1,\dotsc,n$}$$ By defining $w_r = rw_1$ and $b_r=rb_1$, $$\text{Minimize}\quad \|w_r\|=r\|w_1\|\quad\text{s.t.}\quad y_i(w_r/r\cdot x_i+b_r/r) \geq 1\; \text{for $i=1,\dotsc,n$}$$ which is the same as the program: $$\text{Minimize}\quad \|w_1\|\quad\text{s.t.}\quad y_i(w_1\cdot x_i+b_1) \geq 1\; \text{for $i=1,\dotsc,n$}$$

So, though the first and last minimization problems produce different normal vectors and biases, $w_r$, $w_1$, $b_r$, $b_1$; the width of margin remains unchanged, i.e., the margin width is independent of $r$, as:

$$\frac{2r}{\|w_r\|} = \frac{2r}{\|(rw_1)\|}= \frac{2}{\|w_1\|}$$

It depends if you talk about the linearly separable or non-linearly separable case. In the former, the weight vector can be explicitly retrieved and represents the separating hyper-plane between the two classes. The weight associated to each input dimension (predictor) gives information about its relevance for the discrimination of the two classes. In the non-linear case, the hyper-plane is only implicitly defined in a higher dimensional dot-product space by means of the "kernel trick" mapping (e.g. Gaussian kernel replacing the dot product). When using non-linear kernels more sophisticated feature selection techniques are needed for the analysis of the relevance of input predictors.

As I answered to similar question, weight vector of any linear classifier indicates feature importance: simply because final value is a linear combination of feature values with weights as coefficients, so the bigger weight, the more impact to the final value is caused by the corresponding summand.

Thus, for linear classifier you can take features with biggest weights (not with biggest values of the feature itself, or the biggest product of weight and feature value).

It also explains why SVM with non-linear kernels like RBF don't have such a property: both feature values and weights are transformed into another space and you can't say that the bigger weight leads to bigger impact, see wiki.

If you need to select most important features for non-linear SVM, use special methods for feature selection, namely wrapper methods.

Note that even if you are implementing multi-class SVM, the one-vs-all strategy is usually applied (or you can use other strategies to achieve the multi-classifier effect), since SVM is binary classifier.

The dimension of weight vector is the same as feature dimension. So if you are applying the linear SVM, the weight vector dimension is equal to the input space dimension. For other kernels, original data space is transformed into another high-dimensional space, specifically, if you are applying the RBF kernel, the weight vector dimension can be viewed as infinite.

You can search any SVM tutorials online on the topic of the basic binary SVM classifier along with the extended version (such as multi-class SVM, SVR, etc.) . There is also a post on feature space, and a post on interpretation of SVM feature weights that might help. @ogrisel's answer to the 2nd post above gives you a better analysis on one-vs-all strategy on multi-class SVM.