I'm confused about isometries of the flat 2-torus and could't find anything online that cleared my confusion. My problem is the following:
Let $T^2=\mathbb{R}^2/\Gamma$ be a 2-torus for $\Gamma \cong \mathbb{Z}$ a lattice in $\mathbb{R}^2$. I know that any isometry of the euclidean plane which preserves the lattice $\Gamma$ induces an isometry of the flat torus $T^2=\mathbb{R}^2/\Gamma$ and conversly any isometry of the flat torus lifts to an isometry of the euclidean plane which preserves the lattice $\Gamma$.
But my geometric intuition is that any translation in $\mathbb{R}^2$ should descent to an isometry of the flat torus, since translating any "amount" should preserve length of curves on the torus. But when I take for example the standard torus $\mathbb{R}^2/\mathbb{Z}^2$ and a translation $f(x,y)=(x,y+\lambda)$ for $\lambda \not\in \mathbb{Z}$, then for any lattice point $(k,l)$ we have $f(k,l)=(k,l+\lambda)\not\in \mathbb{Z}^2$, hence $f$ does not preserve the lattice $\Gamma$ and hence does not descent to an isometry of the torus. Why is this?
One has to look carefully at what it means for $f$ to "preserve the lattice". I find that phrase to be a little vague and ambiguous, so I'll avoid it, and instead use a more precise phrase. I suspect that if you the source that you are consulting is written with full and precise definitions (and this is a big "if" when using only on-line sources), then it will have the definition that I write, or something equivalent.
Consider a lattice $\Gamma \subset \mathbb R^2$, meaning an additive subgroup which is discrete and isomorphic to $\mathbb Z^2$. We'll say that an isometry $f : \mathbb R \to \mathbb R$ preserves the cosets of $\Gamma$ if for every $(k,l) \in \Gamma$ and every $(x,y) \in \mathbb R^2$ we have $$f(x+k,y+l) - f(x,y) \in \Gamma $$ Equivalently, $f$ preserves the cosets of $\Gamma$ if for every $(k,l) \in \Gamma$ and every $(x,y) \in \mathbb R^2$ there exists $(k',l') \in \Gamma$ such that $$f(x+k,y+l) = f(x,y) + (k',l') $$
This condition is necessary for $f$ to induce an isometry of $T^2$.
In fact, even more is true, this condition it is necessary just for $f$ to induce a well-defined function from $T^2$ to itself.
To see why, fix $(x,y) \in \mathbb R^2$. The point $(x,y)$ is a representative of the coset $(x,y) + \Gamma \in \mathbb R^2 / \Gamma = T^2$. In order for $f$ to induce a well-defined function from $\mathbb R^2 / \Gamma$ to itself, as $(k,l)$ varies over $\Gamma$ it must be true that all of the points $f(x+k,y+l)$ are contained in a single coset of $\Gamma$. That coset contains the point $f(x+0,y+0)=f(x,y)$ (by taking $(k,l)=(0,0)$), and so it follows that for each $(k,l) \in \Gamma$ there exists $(k',l') \in \Gamma$ such that $f(x+k,y+l) = f(x,y) + (k',l')$.
Notice that I have not even talked about isometries, nonetheless I think this addresses your question already: you seem to have overlooked the importance of "preserving cosets".