Answer by paf for Proof or counter-example that $(0,t_0)$ is a maximum of $f: [0,1]^2 \to \mathbb{R}$ if $t_0$ is a maximum of $g(t) := f(0,t)$.

Let's build $f$ such that $g(t) = t-t^2$ (which attains its maximum at $t_0 = \dfrac12$) but which increases with $x$.

For this purpose, let's try $f(x,y) = x + y - y^2$. When we fix a value for $x$, the graph is simply the same parabola but translated to the top by $x$ (see below the drawing in $\Bbb R^3$ of the surface $z = f(x,y)$).

In this case, $f(0,0.5) = g(0.5) = 0.25$ but $f(1,0.5) = 1.25$ so that $(0,0.5)$ isn't a point of maximum of $f$.

Edit: I did not see the condition $D_1f(0,t_0)<0$. Thus I propose a new counterexample $f(x,y) = x^2-\dfrac12 x+y-y^2$ (see the new picture below).

Here, the idea was to add a function of $x$ (that is, $x^2-x$ here) which begins by decreasing when $x$ increases from 0 (in particular, $D_1f(0,0.5) = -\dfrac 12<0$) and then has a sufficiently strong increase when $x$ approaches $1$ so that $f(1,0.5)>f(0,0.5)$.