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Just for completeness, I'll write my own solution from the edit with Wlod AA's notation, in the case where we don't allow the angel to repeat a position (in which case we can't conclude anything about the number of stones, but can conclude something about the size of board needed). As mentioned there is essentially only one thing that can happen in this strategy as well, but it takes twice as long, there are some sidelines, and I need a bounding box that is one cell larger.

Just for completeness, I'll write my own solution from the edit with Wlod AA's notation, in the case where we don't allow the angel to repeat a position (in which case we can't conclude anything about the number of stones, but can conclude something about the size of board needed). 

As mentioned there is essentially only one thing that can happen in this strategy as well, but it takes twice as long as the strategy of Wlod AA, there are some sidelines, and I need a bounding box that is one cell larger. (As mentioned in a comment, one may interpret the difference as follows, in my terminology: Wlod AA skips straight to the pivot part by pretending there is a pivot at $(0, 1)$, and sidelines are skipped by adding a few lemmas.)

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second edit

Just for completeness, I'll write my own solution from the edit with Wlod AA's notation, in the case where we don't allow the angel to repeat a position (in which case we can't conclude anything about the number of stones, but can conclude something about the size of board needed). As mentioned there is essentially only one thing that can happen in this strategy as well, but it takes twice as long, there are some sidelines, and I need a bounding box that is one cell larger.

$$ P_1 := ((1 \; 0) \; \emptyset) $$ $$ P_2 := ((1 \; 0) \; \{(2 \; 0)\}) $$ $$ P_3 := ((1 \; 1) \; \{(2 \; 0)\}) $$ $$ P_4 := ((1 \; 1) \; \{(2 \; 0) \; (1 \; 2)\}) $$ Now we have two choices for the angel. Either they continue running in the cage, or they go right and we begin the pivot process. (As Wlod AA points out, this can be seen as a form of recurrence, but I did not realize this.)

While running around the cage we just drop stones in front of the angel, and this looks like $$ P'_5 := ((0 \; 1) \; \{(2 \; 0) \; (1 \; 2)\}) $$ $$ P'_6 := ((0 \; 1) \; \{(2 \; 0) \; (1 \; 2) \; ({-1} \;1)\}) $$ $$ P'_7 := ((0 \; 0) \; \{(2 \; 0) \; (1 \; 2) \; ({-1} \;1)\}) $$ $$ P'_8 := ((0 \; 0) \; \{(2 \; 0) \; (1 \; 2) \; ({-1} \;1) \; (0 \; {-1})\}) $$ Now you cannot avoid a literal repetition of the position (though indeed we could see the origin as a repetition already). We have to enter a sticky situation.

The angel might as well have entered the sticky situation immediately, and we continue the main line from the fifth step. $$ P_5 := ((2 \; 1) \; \{(2 \; 0) \; (1 \; 2)\}) $$ $$ P_6 := ((2 \; 1) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2)\}) $$

The angel should not go north, as that is a trap (this is Wlod AA's Remark 3): $$ P''_7 := ((2 \; 2) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2)\}) $$ $$ P''_8 := ((2 \; 2) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (2 \; 3)\}) $$ and angel is forced to repeat.

We continue the main line, and repeat the above reasoning for any deviation from the pivot, and get $$ P_7 := ((3 \; 1) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2)\}) $$ $$ P_8 := ((3 \; 1) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \}) $$ $$ P_9 := ((3 \; 0) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \}) $$ $$ P_{10} := ((3 \; 0) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \; (4 \; {-1})\}) $$ $$ P_{11} := ((3 \; {-1}) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \; (4 \; {-1})\}) $$ $$ P_{12} := ((3 \; {-1}) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \; (4 \; -1) \; (3 \; {-1})\}) $$ $$ P_{13} := ((2 \; {-1}) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \; (4 \; {-1}) \; (3 \; {-1})\}) $$ $$ P_{14} := ((2 \; {-1}) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \; (4 \; {-1}) \; (3 \; {-1}) \; (1 \; {-1})\}) $$ $$ P_{15} := ((1 \; {-1}) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \; (4 \; {-1}) \; (3 \; {-1}) \; (1 \; {-1})\}) $$ $$ P_{16} := ((1 \; -1) \; \{(2 \; 0) \; (1 \; 2) \; (3 \; 2) \; (4 \; 1) \; (4 \; {-1}) \; (3 \; {-1}) \; (1 \; {-1}) \; (0 \; {-1})\}) $$ and the angel is forced to repeat.

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Consider a maximal Manhattan distance the Angel can travel, and take a strategy $S$ that takes it the furthest from the initial point against any Devil strategy.

Make a new strategy for the Angel, where at each position, it looks at all possible ways the game can continue, depending on the Devil's strategy played against $S$. Out of all of these, there is an entry to the current node (with at least the same set stones on the table) where the Angel gets the furthest. Now just move in that direction, and pretend there are imaginary stones where the Devil would've put them.

Now the cage argument is trivial, the Angel with the new strategy will run in a circle and after a few moves it reaches a sticky situation, after which it runs around the pivot. Possibly this can be simplified, I did not think about this much, as the OP asked about stone count and I think we can't analyze that after this simplification.

Consider the maximal Manhattan distance the Angel can travel, and take a strategy $S$ that takes it the furthest from the initial point against any Devil strategy.

Make a new strategy for the Angel, where at each position, it looks at all possible ways the game can continue, depending on the Devil's strategy played against $S$. Among all these continuations, and among those the set of all turns where the current node is (re)exited, there is one after which the Angel gets the furthest before reentry (or before being trapped). Now just move directly in that direction, and pretend there are imaginary stones where the Devil would've put them.

Now the cage argument is trivial, the Angel with the new strategy will run in a circle and after a few moves it reaches a sticky situation. The pivot running is also simplified, you can just run around until you reach the end.