The difference is of course that the two penalty terms, $\alpha^{T}K\alpha$ and $\alpha^{T}\alpha$, penalize rather differently. Suppose that $n=m$ is very large and $K(x,y)=K(x-y)$ for some (say even) function $K$ (typical cases should be similar to this).

Then we can consider an infinite-dimensional approximation of this finite-dimensional setting. Let us see how the kernel $K$ acts on the harmonic $e_k$ of frequency $k\in\mathbb R$ given by the formula $e_k(x):=e^{ikx}$ for real $x$:
\begin{equation}
(Ke_k)(x)=\int_{-A}^A K(x-y)e^{iky}dy=\int_{x-A}^{x+A} K(u)e^{ik(x-u)}du\approx\lambda_k e_k(x),
\end{equation}
where $A\in(0,\infty)$ is very large and
\begin{equation}
\lambda_k:=\int_{-\infty}^\infty K(u)e^{-iku}du,
\end{equation}
so that $e_k$ is an approximate eigenvector of $K$ with approximate eigenvalue $\lambda_k$. If $|k|\to\infty$ then, by an appropriate version of the Riemann--Lebesgue lemma, $\lambda_k$ goes to $0$; this convergence is the faster, the smoother $K$ is. So, the RKHS penalizer is lenient with respect to high-frequency harmonics $e_k$, with large $|k|$ -- that is, $e_k^T Ke_k=(Ke_k,e_k)\approx\lambda_k(e_k,e_k)=A\lambda_k$ with $\lambda_k$ small. Accordingly, with the total size $\|\alpha\|_2$ of the minimizing mixture $\alpha$ of harmonics $e_k$ fixed, the RKHS penalty term penalizes mainly the low-frequency constituent harmonics $e_k$ of $\alpha$, with $|k|$ comparatively small. This behavior may result in better catching (by the minimizer) fine, high-frequency features of the unknown, estimated function $f$. However, such behavior may be not so desirable when there is prior knowledge that the true $f$ is rather smooth (whereas some constituent smooth, low-frequency harmonics got partially penalized out).

The ridge penalty term $\alpha^{T}\alpha=(\alpha,\alpha)$ can actually be considered a special case of $\alpha^{T}K\alpha=(K\alpha,\alpha)$ with $K(x-y)=\delta(x-y)$, the delta-function kernel, which is of course very non-smooth. This latter kernel treats all the harmonic frequencies whatsoever absolutely equally: $\delta e_k=e_k$ for all $k$; it is the ultimate "equal opportunity" penalizer, in contrast with the smooth-kernel one.

One should also note that, if $K$ is smooth, the estimate $K\alpha$ of $f$ will to an extent suppress the constituent high-frequency harmonics of $\alpha$, whether $\alpha$ is the RKHS minimizer or the ridge one. However, it should be clear from the above discussion that the overall suppression of the high-frequency harmonics will be relatively less in the RKHS case.