We will first prove that Mari can win in 5 turns. We will denote squares in the grid by the coordinates of their centers, taking Mari's first square to be $(0,0)$. W.l.o.g., let Jüri's first turn be $(1,0)$, allowing him to also color squares with fractional coordinates in future turns (Fig. 7). Mari will color $(2,0)$ on her second move (Fig. 8). W.l.o.g., assume that the $y$-coordinate of Jüri's second move is nonpositive.
The only option for him to color three vertices of some square together with $(1,0)$ and $(0,2)$ is by choosing $(-1,-1)$ (Fig. 9). The only option for him to color three vertices of some square together with $(1,0)$ and $(2,2)$ is by

Fig. 8

Fig. 9

Fig. 10
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choosing $(3, -1)$ (Fig. 10). Thus we consider three cases.

* If Jüri colors $(-1, -1)$ or $(3, -1)$ on his second turn, let it be $(-1, -1)$ without loss of generality (due to symmetry). Then Mari will color $(0, 2)$ (Fig. 11), after which Jüri is forced to color $(2, 2)$ to avoid immediate defeat. Then Mari can color $(-2, 0)$ (Fig. 12), threatening to win by $(-2, 2)$ or $(0, -2)$. Jüri cannot block both squares at once or win himself, so Mari will win on the next move.

* If Jüri colors $(-2, 0)$, $(0, -2)$, $(2, -2)$ or $(4, 0)$, then assume without loss of generality that the $x$-coordinate of this move is greater than $1$. Mari will color $(0, 2)$ again, after which Jüri is forced to color $(2, 2)$ to avoid immediate defeat (the two possible positions are shown in Figures 13 and 14). As $(-2, 2)$, $(-2, 0)$ and $(0, -2)$ are all uncolored, Mari can win exactly like in the previous case.

* If Jüri colors any other square on his second turn, then Mari can color $(0, 2)$ and win just like in the other cases.

Fig. 11
Fig. 12
Fig. 13
Fig. 14

We finally note that Mari cannot win in 4 moves, as after her 3rd move there exists at most 1 square which would yield an immediate victory for her, so Jüri can just block it.