Let's take this in short steps.
"$\Delta{x_i} \sim N(0,k^{-1})$" means that (up to a constant of proportionality which we can worry about at the end) the distribution function (PDF) of $\Delta{x_i}$ equals $k^{1/2}\exp{(-(k/2)\Delta{x_i}^2)}$. This is (one) definition of what it means to have a normal distribution with mean $0$ and variance $1/k$.
"Independent" implies the joint ($n-1$-variate) distribution function of all the $\Delta{x_i} = x_{i+1}-x_{i}$ is the product of the individual distribution functions. Exploit a basic property of the exponential (products of its values correspond to sums of its arguments) to write this joint density as
$$k^{1/2}\exp{(-(k/2)\Delta{x_1}^2)}\cdots k^{1/2}\exp{(-(k/2)\Delta{x_{n-1}}^2)}=k^{(n-1)/2}\exp{\left(-k/2(\Delta{x_1}^2+\cdots+\Delta{x_{n-1}}^2)\right)},$$
once more ignoring the constant.
Finally, because each $x_i$ is obtained by starting with $x_1$ and adding all previous increments,
$$x_i = x_1 + \Delta{x_1} + \Delta{x_2} + \cdots + \Delta{x_{i-1}},$$
we can rewrite the preceding sum of squares $\Delta{x_1}^2+\cdots+\Delta{x_{n-1}}^2$ in terms of the $x_i$ to get $x_1^2+2(x_2^2 + \cdots + x_{n-1}^2)+x_n^2 - 2(x_1x_2 + x_2x_3 + \cdots + x_{n-1}x_n)$. Plug this into the preceding formula to obtain the joint distribution of the $x_i$.
One usually doesn't care much about the details of this resulting (somewhat messy) expression; its form is what matters. Apart from constant multipliers, which will be determined by the integrate-to-unity criterion satisfied by any probability distribution, it is the exponential of a quadratic form $Q$. This means the $x_i$ have a multivariate normal distribution. You can read off their variances and covariances by inverting $Q$.