MST with degree constraints on some node
Here's an algorithm that just reduces to MST; no need to modify Prim's or some other algorithm. The idea is simple: remove $v$, compute the MSTs of the resulting components of the graph, and then stitch them together with $v$. The interesting case is when removing $v$ doesn't disconnect the graph.

I'll assume the input graph $G$ is connected, although it would not be difficult to generalize this idea to computing a minimum spanning forest. Also, obviously the degree of $v$ needs to be at least 2.

Algorithm

Start by removing $v$ and all incident edges; call this $G'$. Now compute the number of connected components in $G'$. This can be done with DFS or something similar.

More than 2 components:
If $G'$ has more than 2 connected components, no solution is possible.

Exactly 2 components: If $G'$ has exactly two connected components, then we can construct a solution by using $v$ as the "bridge" between the two components: just connect the MSTs of the components with the cheapest edges through $v$.

Exactly 1 component: If $G'$ has exactly one component, we can compute $T' = \text{MST}(G')$, and then we just need to stitch $v$ into $T'$ using two edges.


Find the cheapest edge incident on $v$ and add this to $T'$. Call this $T''$. Now we have a spanning tree, but not necessarily minimum.
Iterate over all other edges incident on $v$. For each of these, we will try to construct a MST where $d(v) = 2$. If we don't find one, then no solution is possible.


Consider some incident edge $(u,v)$ which is not the lightest edge on $v$. 
Observe that if we added this edge to $T''$, it would create a cycle.
If $(u,v)$ is the heaviest edge on this cycle, then skip this edge, because no solution is possible using $(u,v)$.
Otherwise, we can take out the heaviest cycle edge and add $(u,v)$ instead. Now we have a spanning tree where $d(v) = 2$. If this spanning tree is minimum, we're done. (We can determine if a spanning tree is minimum just by computing any $\text{MST}(G)$ and comparing weights. This "reference" MST only needs to be computed once.)



It's not immediately obvious that this is within cost bounds, because for each of $O(|V|)$ edges incident on $v$, we have to compute the heaviest edge in the tree on the path between $v$ and some other vertex $u$. However, these "path queries" are well studied in the literature, and can be computed in $O(\log|V|)$ by preprocessing $T''$ into a balanced lookup structure. Some examples include binary lifting, Miller-Reif rake/compress tree contraction, and Sleator-Tarjan link/cut trees. The link/cut trees are actually far more general: they handle dynamic trees too! But here we only need to preprocess a static tree, i.e. $T''$.



Edit: My original idea was to do the following, which doesn't work (thank you @VladislavBezhentsev for pointing this out in the comments!). It doesn't work because, while we know that the cheaper of $e_1$ and $e_2$ is certainly in the MST (by the cut property), it's not necessarily true that the heavier of the two edges is!


  (Incorrect idea): Find the two cheapest edges $e_1$ and $e_2$ that are incident on $v$. Add $e_1$ and $e_2$ to $T'$: this creates a cycle. If the largest edge on this cycle is either $e_1$ or $e_2$, no solution is possible. Otherwise, remove the largest edge on the cycle.