There's a matrix identity
$$
PB^T(BPB^T + R)^{-1} = (P^{-1} + B^TR^{-1}B)^{-1}B^TR^{-1}
$$
for matrices with concordant dimensions and invertibility.

Taking $P = I$, $R = \lambda I$, $B^T = H$, and $B = H^T$, we have
$$
H(H^TH + \lambda I)^{-1} = (\lambda I + HH^T)^{-1}H.
$$
Let
$$
h_x = \left(\begin{array}{c}h_1(x) \\ \vdots \\ h_M(x)\end{array}\right)
$$
so we have
$$
\hat f(x) = h_x^T\hat\beta = h_x^T(H^TH + \lambda I)^{-1}H^Ty \\
= h_x^TH^T(HH^T + \lambda I)^{-1}y \\
= \hat\alpha^T(Hh_x)
$$
by the aforementioned identity. Now 
$$
(Hh_x)_i = \langle h_{x_i}, h_x\rangle = K(x, x_i)
$$
so all together we have
$$
\hat f(x) = \sum_{i=1}^N \hat\alpha_i K(x,x_i)
$$

-------

To prove that identity, you can note that
$$
PB^T(BPB^T + R)^{-1} - (P^{-1} + B^TR^{-1}B)^{-1}B^TR^{-1} \\
= (P^{-1} + B^TR^{-1}B)^{-1}\left[(P^{-1} + B^TR^{-1}B)PB^T - B^TR^{-1}(BPB^T + R)\right](BPB^T + R)^{-1} \\
= C\left[B^T + B^TR^{-1}BPB^T - B^TR^{-1}BPB^T - B^T\right]D \\
= \mathbf 0
$$