Let $a>0$, $f\in C^\infty(\mathbb{R}^d)$ and $f_a\in C^\infty(\mathbb{R}^d)$ defined by $$ f_a(x):= f\left(\frac{x}{a}\right). $$ Denoting by $B_s$ the ball of center $0\in \mathbb{R}^d$ and radius $s$.
Is it the $L^\infty$ norm of $\Delta f_a$ on $B_a$ equal to $a^{-2}\| \Delta f \|_{L^\infty (B_1) }$ or $\| \Delta f \|_{L^\infty (B_1) }$ ?
More precisely, $$ \| \Delta f_a \|_{L^\infty (B_a) }= \left\{ \begin{aligned} a^{-2}\| \Delta f \|_{L^\infty (B_1) } \\[4pt] or \\[4pt] \| \Delta f \|_{L^\infty (B_1) } \end{aligned} \right. \qquad ? $$
My confusion become from how can I handle the $\Delta$ operator.
I tried the following: $$ \Delta f_a (x) = \Delta f\left(\frac{x}{a}\right),$$ so doing the change of variable $u=\frac{x}{a}$ I got $$ \Delta f\left(\frac{x}{a}\right) = \Delta f(u),$$ and considering that if $x=a$ then $u=1$, and then I concluded $$ \| \Delta f_a \|_{L^\infty (B_a) } = \| \Delta f \|_{L^\infty (B_1) }. $$
Is it correct ?
There is a mistake in your computation when you write $\Delta f_a(x) = \Delta f(\frac x a)$, which stems from incorrect parenthesis / priorization of operations.
Let us go slowly. You start from you function $f$, from which you define a new function $f_a$, which occurs to be defined by $f_a(x) = f(\frac x a)$. Now you want to compute the Laplacian of $f_a$, $\Delta f_a$ and evaluate it at some point $x$, so you want $(\Delta f_a)(x)$. As a side note, computing $\Delta (f_a(x))$ would not really make sense since $f_a(x)$ is a (fixed) real number, and not a function.
Now, let us compute $\Delta f_a = \partial_{x_1}^2 f_a + \dotsb + \partial_{x_d}^2 f_a$. One way to make things clear is that $f_a = f \circ \sigma$ where $\sigma(x) = \frac{x}{a}$. By the chain rule $$\partial_{x_i} f_a = \partial_{x_i} (f \circ \sigma) = (\partial_{x_i} \sigma) ((\partial_{x_i} f) \circ \sigma) = \frac 1 a (\partial_{x_i} f) \circ \sigma $$ Differentiating once more and summing over $i$, you obtain $\Delta f_a = \frac{1}{a^2} (\Delta f) \circ \sigma$. So the correct version is indeed $$ \| \Delta f_a \|_{L^\infty(B_a)} = \frac 1 {a^2} \| \Delta f \|_{L^\infty(B_1)}. $$
Heuristically, when $a$ is small, you are "concentrating" your function. Such scalings make the derivatives larger, since the function must vary of the same amplitude but on shorter scales. So it is normal to have a $a^{-2}$ factor here for second-order derivatives.