Why are the random variables $X+Y$ and $X-Y$ independent when $X$ and $Y$ are i.i.d $N(0,1)$?
(1) The short, short answer is that it is wrong to say
$$\mathbb{P}(X+Y=u|X-Y=v)\neq \mathbb{P}(X+Y=u)\,\,\,\,\,\,\text{(this is wrong)}$$
because in fact, both sides $=0$, as these are continuous variables.
(2) The longer answer... Well first of all, the proper way to decide independence is to look at the joint PDF of $U = X+Y$ and $V=X-Y$, as you have already done. This is equivalent to checking:
$$f_U(U = u) \overset{?}= f_{U|V}(U = u \mid V = v) \equiv \frac{f_{U,V}(U = u \cap V = v)}{f_V(V = v)}$$
where you will find that both sides are non-zero and indeed equal.
(3) However, I wonder if your confusion comes from a more basic misunderstanding. It is of course true that $(U,V) = (u,v)$ defines exactly a single point in $(X,Y)$ space. However this does not automatically imply the conditional (prob or density) is $<$ the unconditional. After all, remember that all conditional prob (or density) are ratios. So if the numerator is very small but the denominator is proportionally small, then the ratio is unchanged and the conditional prob (or density) equals the unconditional version.
In your example, the unconditional asks for hitting a certain line $X+Y = u$ within the entire $2$-D $(X,Y)$ plane, while the conditional asks for hitting a point within a specific line $X-Y = v$. As mentioned, both probabilities are zero, but as you verified, both densities are non-zero and equal.
(4) Finally, you might like to know that multivariate Gaussians are the only variables with this property. So that might explain why your gut just keeps telling you that $X+Y, X-Y$ "cannot possibly be independent" when $X,Y$ are independent. :) I was confused about this in the recent past -- see this for a brief further discussion.
To understand a very intuitive brainstorming let's start with $X,Y$ iid $N(\theta;1)$ distribution.
You will probabily know that $X+Y$ is a "complete sufficient statistic" for $\theta$ while $X-Y\sim N(0;2)$ is independent of $\theta$ so it is "ancillary"
This is that $X+Y$ contains all the information about $\theta$ while $X-Y$ has no useful information...its distribution does not depend anymore from $\theta$
So they are independent
This intuitive brainstorming is, in poor words, Basu's Theorem