The question is why:
$$ \|f \|^2_{\mathcal H} = \langle c , Kc \rangle_{R^n} $$
I will be very thorough.
The original space $\mathcal H $ is an RKHS which means that it has an norm induced by its inner product:
$$ \| f \|^2_{\mathcal H} = \langle f, f \rangle_{\mathcal H}$$
by the representer theorem we know:
$$ \min_{f \in \mathcal H} \frac{1}{n} \sum^n_{i=1} L(f(x_i),y_i) + \lambda \|f \|^2_{\mathcal H} = \min_{f \in \mathcal{ \hat{H} }} \frac{1}{n} \sum^n_{i=1} L(f(x_i),y_i) + \lambda \|f \|^2_{\mathcal H} $$
therefore we only need to consider functions in $\mathcal{ \hat{H} }$. This means that we only care about function that depend on the data and have form $f(\cdot) = \sum^n_{i=1} c_i K(x_i, \cdot) = f = \sum^n_{i=1}c_i K_{x_i}$. Therefore since we are only considering functions of that form we can plug this in the definition of the norm and we get:
$$ \| f \|^2_{\mathcal H} = \langle f, f \rangle_{\mathcal H} = \langle \sum^n_{i=1}c_i K_{x_i}, \sum^n_{i=1}c_i K_{x_i} \rangle_{\mathcal H}$$
recall inner products are linear function (i.e. $\langle f + g, h \rangle_{\mathcal H} = \langle f, h \rangle_{\mathcal H} + \langle g, h \rangle_{\mathcal H}$ or similarly $f(\sum_j x_j, \cdot) = \sum_j f(x_j, \cdot)$). Thus we can apply the function $\langle \cdot, \cdot \rangle$ to each argument (i.e. to each function $K_{x_i}$ individually):
$$ \langle \sum^n_{i=1}c_i K_{x_i}, \sum^n_{i=1}c_i K_{x_i} \rangle_{\mathcal H} = \sum^n_{i=1}c_i \langle K_{x_i}, \sum^n_{i=1}c_i K_{x_i} \rangle_{\mathcal H} $$
and again (apply linearity):
$$ \sum^n_{i=1}c_i \langle K_{x_i}, \sum^n_{i=1}c_i K_{x_i} \rangle_{\mathcal H} = \sum^n_{i=1} \sum^n_{j=1}c_i c_j \langle K_{x_i}, K_{x_j} \rangle_{\mathcal H} $$
now here is the interesting part where we use the fact we are in a RKHS. Recall an RKHS has the so called "reproducing property". i.e. its continuous evaluation functionals lead it to have the remarkably property:
$$ \langle f, K_{x} \rangle_{\mathcal H} = f(x) $$
whenever $f$ and $K_x$ are in the RKHS. In this case it happens that $K_{x_i} = K(x_i, \cdot) \in \mathcal H$. Which means we can use the reproducing property:
$$ \langle f, K_{x_i} \rangle_{\mathcal H} = f(x_i) $$
Thus using the reproducing property (what people call the "kernel trick", i.e. we don't need explicit feature maps, just the inner products in this case $K(x_i, x_j)$) we get:
$$ \langle K_{x_i}, K_{x_j} \rangle_{\mathcal H} = K_{x_i}(x_j) = K(x_i, x_j) $$
Now its not hard to see that you can organize the $K(x_i, x_i)$ into a matrix and you get:
$$ \sum^n_{i=1} \sum^n_{j=1}c_i c_j \langle K_{x_i}, K_{x_j} \rangle_{\mathcal H} = \langle c, Kx \rangle_{R^n} = c^T K c$$
or you can just recall from linear algebra or some other course (maybe optimization) that $$ \sum^n_{i=1} \sum^n_{j=1}c_i c_j \langle K_{x_i}, K_{x_j} \rangle_{\mathcal H} $$ is simply a quadratic form and can be expressed as $c^T K c$.
The important bit to notice about this proof is that we started off with an infinite dimensional space and we ended up being to compute its norm in a finite dimensional world! Geesh, thanks representer theorem! ;)
Notice that there is one last point I don't know how to answer and why the linear case does not have that inner product. If someone else can clarify that last bit this question is essentially done! :)