Need help applying Kruskal's Algorithm to an existing adjacency matrix program that uses a 2D struct to store the data

At first I want to make a note: in fact, your node array describes edges, not nodes. Nodes here are indexes. Anyway, I leave the name as is. I presume that your graph is undirected. Here is how Kruskal algorithm can be implemented with your structure.

Define the function:

 std::vector<std::pair<int, int>> kruskal() 
 {
     std::vector<std::pair<int, int>> mst; //our result

At the very beginning we split all the vertices to separate trees. Each tree is identified by an index. We create a lookup table treesByVertex for finding an index of a tree by vertex.

    std::map<int, std::set<int>> trees;
    std::map<int, int> treeByVertex;
    for (int i = 0; i < MAXNODES; ++i)
    {
        std::set<int> tree; // a tree containing a single vertex
        tree.emplace(i); 
        trees.emplace(i, tree); //at startup, the index of a tree is equaled to the index of a vertex
        treeByVertex.emplace(i, i);
    }

Then we create a helper structure edges that will contain a list of edges with ascending distances:

    std::multimap<int, std::pair<int, int>> edges;
    for (int i = 1; i < MAXNODES; ++i)
        for (int j = 0; j < i; ++j)
            if (node[i][j].exists)
                edges.emplace(node[i][j].distance, std::make_pair(i, j));

Iterate over all the edges in ascending order and test if it connects two different trees. If it's true, we add this edge to mst and merge these two trees:

    for (const auto& e : edges)
    {
        int v1 = e.second.first;
        int v2 = e.second.second;
        if (treeByVertex[v1] != treeByVertex[v2]) //use our lookup table to find out if two vertexes belong to different trees
        {
            mst.emplace_back(v1, v2); //the edge is in mst
            trees[v1].insert(trees[v2].begin(), trees[v2].end()); //merge trees
            for (int v : trees[v2]) //modify lookup table after merging
                treeByVertex[v] = treeByVertex[v1];
        }
    }

    return mst;
}

In fact, you even don't need trees container here at all.