k-Vertex Cover problem is in parameterized Log space
Here is an algorithm that uses $2k^2 + O(\log n)$ space. This is just the observation that the well known "Buss kernel" for Vertex Cover can be computed in log-space:

Say that a vertex has big degree if it has degree at least $k+1$. All vertices of big degree must be in every vertex cover of size at most $k$. If $v$ does not have big degree it has small degree. Call a vertex $v$ interesting if it has small degree and has at least one neighbor of small degree. Any inclusion minimal vertex cover of $G$ of size at most $k$ must contain all vertices of big degree, and out of the vertices of small degree, it can only contain interesting vertices.

One can easily verify that in $O(\log n)$ space one can determine whether a given vertex has small or big degree, and whether it is interesting or not. Further, we can count the number of big degree / small degree / intersting vertices before $v$ in input. Thus, given an integer $j$ we can also find the $j$'th vertex of big degree / small degree or the $j$'th interesting vertex.

The algorithm computes the number $b$ of vertices of big degree, if $b &gt; k$ the algorithm outputs that there is no vertex cover of size at most $k$. The algorithm computes the number $i$ of interesting vertices. If $i &gt; 2k^2$ it outputs that there is no vertex cover of size at most $k$ (because one has to cover at more than $k^2$ edges using only vertices of degree at most $k$).

Now the algorithm goes over all the $2^i$ possible binary strings on length $i$. For each such string, we interpret the $j$'th bit as whether the $j$'th interesting vertex is in the vertex cover or not. 

For each string compute the total size of the proposed solution ($b$ + the number of ones in the binary string), and check whether the proposed solution is a vertex cover by going over every pair of interesting vertices, checking whether they are adjacent, and whether both are $0$ in the binary string. This concludes the algorithm.

Remark: because the number of interesting vertices is at most $O(k^2)$ and we can determine whether a given vertex is interesting or not, one could run any vertex cover algorithm on the subgraph induced on the interesting vertices (for example the $O(1.2738^k + nk)$ time algorithm by Chen, Kanj and Xia), and this would still only use $f(k) + O(\log n)$ space.