In short, because — as emphasized by the last phrase of your bolded passage — having extra pieces on the board in Tic-Tac-Toe is never bad. It's not just 'having more squares' vs. 'having fewer squares'; it's that if position A is exactly position B with an extra X on it, then position A is always at least as good for player X as position B is.
So now suppose you're X, the first player, and you're strategy-stealing; and suppose you come across a moment where the square you're 'supposed' to make your move in — vs. the opponent's given plays — is already taken. Then if that square weren't filled, you would be moving to fill it, meaning that you're moving to some position $P_0$ where (by the assumptions) you're guaranteed to have a winning position. But instead, because you already hold the square you're putting your X somewhere random on the board instead. Then that position is simply $P_0$+X. By the argument in the previous paragraph, this is at least as good for you as position $P_0$ is; but since we knew (by strategy) that $P_0$ was a winning position for you, then the new position $P_0$+X is winning too.
I will quote some results and problems from the book Combinatorial Games: Tic-Tac-Toe Theory by József Beck, some of which were also quoted in this answer.
The terms "win" and "draw" refer to the game as ordinarily played, i.e., the first player to complete a line wins. The term "Weak Win" refers to the corresponding Maker-Breaker game, where the first player ("Maker") wins if he completes a line, regardless of whether the second player has previously completed a line; in other words, the second player ("Breaker") can only defend by blocking the first player, he cannot "counterattack" by threatening to make his own line. (Note that ordinary $3\times3$ tic-tac-toe is a Weak Win.) A game is a "Strong Draw" if it is not a Weak Win, i.e., if the second player ("Breaker") can prevent the first player from completing a line.
Theorem 3.1 Ordinary $3^2$ Tic-Tac-Toe is a draw but not a Strong Draw.
Theorem 3.2 The $4^2$ game is a Strong Draw, but not a Pairing Strategy Draw (because the second player cannot force a draw by a single pairing strategy.)
Theorem 3.3 The $n\times n$ Tic-Tac-Toe is a Pairing Strategy Draw for every $n\ge5.$
For a discussion of Oren Patashnik's computer-assisted result that $4^3$ tic-tac-toe is a first player win, Beck refers to Patashnik's paper:
Oren Patashnik, Qubic: $4\times4\times4$ Tic-Tac-Toe, Mathematics Magazine 53 (1980), 202-216.
Not much more is known about multidimensional tic-tac-toe, as can be seen from the open problems:
Open Problem 3.2 Is it true that $5^3$ Tic-Tac-Toe is a draw game? Is it true that $5^4$ Tic-Tac-Toe is a first player win?
The conjecture that "if there is a winning strategy on $a^d$, there is one also on $a^{d'}$ for any $d'\geq d$" is given as an open problem:
Open Problem 5.2 Is it true that, if the $n^d$ Tic-Tac-Toe is a first player win, then the $n^D$ game, where $D\gt d$, is also a win?
Open Problem 5.3. Is it true that, if the $n^d$ game is a draw, then the $(n+1)^d$ game is also a draw?
To see that the intuition "adding more ways to win can't turn a winnable game into a draw game" is wrong, consider the following example of a tic-tac-toe-like game, attributed to Sujith Vijay: The board is the set $V=\{1,2,3,4,5,6,7,8,9\};\ $ the winning sets are $\{1,2,3\},$ $\{1,2,4\},$ $\{1,2,5\},$ $\{1,3,4\},$ $\{1,5,6\},$ $\{3,5,7\},$ $\{2,4,8\},$ $\{2,6,9\}$. As in tic-tac-toe, the two players take turns choosing (previously unchosen) elements of $V;$ the game is won by the first player to succeed in choosing all the elements of a winning set. It can be verified that this is a draw game, but the restriction to the board $\{1,2,3,4,5,6,7\}$ (with winning sets $\{1,2,3\},$ $\{1,2,4\},$ $\{1,2,5\},$ $\{1,3,4\},$ $\{1,5,6\},$ $\{3,5,7\}$) is a first-player win.
Best Answer
The paper by Develin and Payne referenced in the comments gives a very thorough treatment of this type of game, and in particular addresses the complicated nuances of breaking ties that arise during the bidding. However, in the Richman games they treat, the winning bid is given to the other player... this problem is slightly different. Glossing over the nuances of tied bids, we can see the following. (Assume that $B$ has one unit to bid with, and that both players will bid and play optimally.)
We see two constraints on a winning bid for player $A$. First, $y \le x - x(G_A)$ for some state $G_A$ that player $A$ can move to; i.e., $y \le x - \min_{G_A} x(G_A)$. Second, either $y \ge 1$ (so player $B$ can't outbid it), or else $y\ge 1 - x / x(G_B)$ for all states $G_B$ that player $B$ can move to; i.e., $y \ge 1 - x / \max_{G_B} x(G_B)$. The latter constraint is never stronger than the former, so we have: $$ 1-x/\max_{G_B}x(G_B) \le y \le x-\min_{G_A}x(G_A). $$ This interval of winning bids shrinks as $x$ decreases, and vanishes (by definition) when $x=x(G)$. So setting the two endpoints equal to one another lets us solve for $x(G)$: $$ 1-x(G)/\max_{G_B}x(G_B) = x(G)-\min_{G_A}x(G_A) \implies x(G)=\frac{1+\min_{G_A}x(G_A)}{1+1/\max_{G_B}x(G_B)}. $$
At this point we can calculate. I'm pasting Python code to evaluate this recursion below. The result that I get for Tic-Tac-Toe is $x(G_0)\approx 1.018375$. I have modified the code to give an exact (rational) result (code not shown here, as it's not worth the extra space): it is $$x(G_0)=\frac{396925}{389763}=1+\frac{7162}{389763}.$$