I assume that isomorphic embedding really meant isometric embedding in the assumptions and thus $Y$ is really a convex subset of $X_i$ for all $i$.

Then the CAT(0)-space on which the amalgamated product acts may be thought of as a thickened version of the Bass-Serre tree.

Usually the Bass-Serre tree has as vertex set $G/F_0 \amalg G/F_1 = G\times_{F_0}pt \amalg G\times_{F_1} pt$ and as edges $G/H = G\times_{H} pt$.

Then the whole tree is constructed by identifying $(gH,0)\in G/H\times [0,1]$ with its coset in $gF_0\in G/F_0$ and $(gH,1)\in G/H\times [0,1]$ with its coset in $gF_1\in G/F_1$ (using the inclusions $H\to F_i$).

Now the big space on which the amalgamated product acts can be constructed similarly. The idea is to thicken the points, i.e. take $G\times_{F_0}X_0\amalg G\times_{F_1}X_1\amalg G\times_{H}Y\times [0,1]$ subject to the identifications $(g,y)\sim (g,f_i(y))$ for $i=1,2$.

Now there are a couple of questions:

1. Is that space still contractible after forgetting the $G$-action. Yes (one can also use this contruction to build a model for $E(F_0\ast_HF_1)$ out of models for $EF_i$ and $EH$ without the metric assumptions and embeddings.

2. Is it still locally CAT(0) ? Yes, this should be the usual glueing lemma for CAT(0)-spaces along convex subspaces.

Thus the result is (globally) CAT(0). If $Y$ and each $X_i$ is a CAT(0) cube complex and the inclusions of $Y\to X_i$ are inclusions of $CAT(0)$-cube complexes, the whole construction should also give a CAT(0)-cube complex.

The same idea also works for HNN-Extensions, although in this case it might be easy to forget to check all assumptions (like for example for $NS(1,2)$).

I assume that isomorphic embedding really meant isometric embedding in the assumptions and thus $Y$ is really a convex subset of $X_i$ for all $i$.

Then the CAT(0)-space on which the amalgamated product acts may be thought of as a thickened version of the Bass-Serre tree.

Usually the Bass-Serre tree has as vertex set $G/F_0 \amalg G/F_1 = G\times_{F_0}pt \amalg G\times_{F_1} pt$ and as edges $G/H = G\times_{H} pt$.

Then the whole tree is constructed by identifying $(gH,0)\in G/H\times [0,1]$ with its coset in $gF_0\in G/F_0$ and $(gH,1)\in G/H\times [0,1]$ with its coset in $gF_1\in G/F_1$ (using the inclusions $H\to F_i$).

Now the big space on which the amalgamated product acts can be constructed similarly. The idea is to thicken the points, i.e. take $G\times_{F_0}X_0\amalg G\times_{F_1}X_1\amalg G\times_{H}Y\times [0,1]$ subject to the identifications $(g,y)\sim (g,f_i(y))$ for $i=1,2$.

Now there are a couple of questions:

1. Is that space still contractible after forgetting the $G$-action. Yes (one can also use this contruction to build a model for $E(F_0\ast_HF_1)$ out of models for $EF_i$ and $EH$ without the metric assumptions and embeddings.

2. Is it still locally CAT(0) ? Yes, this should be the usual glueing lemma for CAT(0)-spaces along convex subspaces.

Thus the result is (globally) CAT(0). If $Y$ and each $X_i$ is a CAT(0) cube complex and the inclusions of $Y\to X_i$ are inclusions of $CAT(0)$-cube complexes, the whole construction should also give a CAT(0)-cube complex.

The same idea also works for HNN-Extensions, although in this case it might be easy to forget to check all assumptions (like for example for $BS(1,2)$).

Let me just stress how crucial the assumption that each $Y\to X_i$ is an isometric embedding really is. For example, if one drops this assumption it is in general unknown whether the result is a CAT(0) group. Usually this construction does not give a CAT0 space but sometimes one can find a better space. For example for free by cyclic groups I asked this question a long time ago When is a extension of $\mathbb{Z}$ by a free group a CAT(0) group?.