CSES - Even Outdegree Edges | Upward Backedge Proof or Intuition?

Although i understood the solution given in usaco guide but still i cannot convince myself why the following statement always produce correct result?
Direct all edges that aren't part of this tree "upwards".
i’ve already read the cf blog given in guide but still cannot get intuition behind this.
Any help would be appreciated.

relevent code

void dfs(int u,int p){
    vis[u] = timer++;
        if(v==p) continue;
            if(vis[v] <= vis[u]){ // check if it is backedge that is going from child to its ancestor
                odd[u] =odd[u]^1;
                ans.push_back(mp(u,v)); // assign upward edge why?
            if(odd[v]) {ans.push_back(mp(v,u)); odd[v]=0;}
            else{ ans.push_back(mp(u,v)); odd[u] =odd[u]^1;}


Did you ever understand this? CP is really pushing the limits of conscious lol

Here’s a different (but ultimately similar) way to think about a solution to the problem.
Imagine you are solving a completely different problem, given a tree with vertices labeled either 0 or 1. You must decide the directions of each edge to make all vertices equal to 0. Like the original problem, setting an edge away from a vertex toggles its value.

In this simple problem, there is a very simple greedy solution. If a leaf has a value of 1, the edge adjacent to it must be an out edge to make it 0. Otherwise, it must be an inward facing edge because making it an out edge would make it 1 and there are no other edges incident to it to fix it. After you have resolved a leaf, remove it from the graph, and if the edge is an in edge change the value of the parent.

Eventually you will end up with a single node after applying the greedy algorithm for many steps. The value of the node is 0, then our algorithm found a correct configuration.

How do we know in the case that our algorithm fails to find a solution there are none? Denote the sum of the nodes initial states A and the number of edges M. All we care about is the parity of A. Note that every time we assign an edge we add 1 mod 2 to a node changing the parity of A. Hence the final parity of A is A + M mod 2. Obviously if this is odd then there are no solutions. Therefore the case where we end up with one odd node A was odd from the beginning, meaning there were no solutions for our greedy algorithm to find.

Now how does this simplified problem help us with the one at hand? Well, we have demonstrated that we can always solve it on any arbitrary tree with nodes preset to degrees of either even (0) or odd (1). Therefore it doesn’t matter which trees we choose as long as they cover as many nodes as possible. We can create an arbitrary set of spanning trees in our graph to make one to run our algorithm on by finding them and setting all edges not in the tree arbitrarily.

DFS in the model solution is simply an easy way of constructing these trees. A more traditional algorithm using DSU can alternatively be used to construct these if one prefers.