Given four rational numbers $m, n, p$, and $q$, define the lattices $L_1 = \{ (m\cdot x, n\cdot x) : x \in \mathbb{Z} \}$ and $L_2 = \{ (p\cdot y, q\cdot y) : y \in \mathbb{Z} \}$. That is, the lattices corresponding to the integer multiples of the fixed vectors $(m, n)$ and $(p, q)$.
Define the a lattice $L = L_1 + L_2 = \{ (m\cdot x + p\cdot y, n\cdot x + q\cdot y) : x, y \in \mathbb{Z} \}\subset \mathbb{Q}^2$.
Then let $L^*$ be the projection of $L_2$ orthogonally to the rational span of $L_1$ and $\lambda := \lambda_1(L^*)$ to be the length of the shortest non-zero vector of $L^*$.
Does the following hold? $$\forall u \in L, ||u|| < \lambda \Rightarrow u \in L_1$$
(Obs: this question is a simplification of this one.)
If by "orthogonal" you mean a projection orthogonal to $L_2$, then unless $(m,n)$ and $(p,q)$ are parallel, $$\forall u \in L, ||u|| < \lambda \implies u \notin L_1$$
Let $\vec m = (m,n), \vec p = (p,q)$, and let $\theta$ be the angle between $L_1$ and $L_2$, so, $\vec m \cdot \vec p = \|\vec m\|\|\vec p\|\cos \theta$.
The length of the orthogonal projection of an element $t\,\vec m \in L_1$ onto $L_2$ will be $|\cos \theta|$ times the length of the original element. I.e., $$|t|\|\vec m\||\cos\theta|$$ As $\|\vec m\|$ and $\theta$ are constant, the only way to decrease this is to decrease $|t|$, and thus the length of the smallest non-zero vector in $L^*$ must be $\lambda = \|\vec m\||\cos\theta| < \|\vec m\|$. But every vector in $L_1$ is an integral multiple of $\vec m$, and therefore has norm $\ge \|m\|$.
Therefore any vector $u$ (in $L$ or not) with $\|u\| < \lambda$ cannot be in $L_1$.
If instead you mean a projection orthogonal to $L_1$, then the norm of the projection of $t\,\vec m$ onto $L_2$ will be $$|t|\|\vec m\|\sec \theta$$ and now $\lambda = \|\vec m\||\sec \theta|$
When $\|\vec p\| < \|\vec m\|$, clearly $u = \vec p$ is a counter-example to the claim. Possibly, there are some $(m,n)$ and $(p,q)$ for which the claim can hold, but it is certainly not true in general.