It appears that the matrix from rbf is a correlation matrix, with ones on the diagonal.  It's also a real-square symmetric matrix, since the upper triangular (off-diagonal elements) are a mirror image of the lower triangular.

The cov matrix looks like a covariance matrix, with diagonal elements that are the variance of each column, and off-diagonals are covariances.  You can check if the following are true:

$\sigma_{1}^2 = \frac{\sum_{i=1}^3 (x_{i1} - \bar{x}_1 )^2}{n-1} = 0.1803    \rightarrow\textrm{variance of feature 1}$

$\sigma_{12} = \frac{\sum_{i=1}^3 (x_{i1} - \bar{x}_1 )(x_{i2} - \bar{x}_2 )}{n-1} = 0.4715   \rightarrow\textrm{covariance between feature 1 and 2}$

If so, the cov is the covariance matrix.

I would not commingle kernel methods with the covariance matrix, mostly because kernel tricks could throw you off.  If you stick with statistical notation and calculations for obtaining the covariance matrix: 

\begin{equation}
\boldsymbol{\sigma} = 
\begin{bmatrix}
\sigma_1^2       & \sigma_{12}    & \cdots  & \sigma_{14} \\
\sigma_{21}   &  \sigma_2^2    & \cdots  &  \sigma_{24} \\
\vdots  & \vdots  & \ddots  & \vdots  \\
\sigma_{41} & \sigma_{42}  & \cdots  &  \sigma_4^2  \\
\end{bmatrix},
\end{equation}

then you can't go wrong.   The correlation matrix is simply:

\begin{equation}
\boldsymbol{\rho}  = 
\begin{bmatrix}
1       & \rho_{12}    & \cdots  & \rho_{14} \\
\rho_{21}   &  1    & \cdots  &  \rho_{24} \\
\vdots  & \vdots  & \ddots  & \vdots  \\
\rho_{41} & \rho_{42}  & \cdots  &  1 \\
\end{bmatrix},
\end{equation}

where all diagonal elements are ones, and elements in the off-diagonal are calculated as 

$\rho_{jk} = \frac{\sigma_{jk}}{\sigma_j \sigma_j}$,

where $\sigma_j $ = $\sqrt{\sigma_j^2}$, which are on the diagonal of the covariance matrix.