Say you have the square pulse function
$$f(x) = \left\{\begin{array}{ll}1 &\text{for $-a \leq x \leq a$}\\ 0&\text{otherwise} \end{array}\right.$$
When calculating the fourier transform I get
$$ f(\omega) = 2\frac{\sin(a\omega)}\omega.$$
I know that the fourier transform is used to show what sinusoidal components a function is made up of. But I cannot quite see it from this result.
How does $f(\omega)$ show me what $f(x)$ is composed of? What is the meaning of $f(w)$?
This is a difficult question because "viewing" in the frequency domain is somewhat unintuitive. One can only give some landmarks to build an intuition of the meaning of the Fourier transform, so my answer will not be very rigorous.
I believe you are aware of the correspondence between sinusoidal functions like $x\mapsto\cos(\omega_0 x)$ and the Dirac delta-function $\omega\mapsto\delta(\omega-\omega_0)$. If I note the complex exponential as $\operatorname{expi}:x\mapsto\exp(\mathrm i x)$, I will denote the correspondence in the symbolic form
$$ \operatorname{expi}\underset{\mathcal F}\longleftrightarrow \delta. \tag{1}$$
This symbolic form is symmetric, I could have also written $\delta\underset{\mathcal F}\longleftrightarrow\operatorname{expi}$ to express exactly the same thing.
There is another remarkable Fourier pair involving the Gaussian function $g(x)=\exp(-x^2/2)$. The Fourier transform of $x\mapsto \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{x^2}{2\sigma^2}\right)$ is $\omega\mapsto\exp\left(-\frac{\sigma^2\omega^2}{2}\right)$.
$$g\underset{\mathcal F}\longleftrightarrow g\tag{2}$$
Now consider your gate function $f$ and its Fourier transform that I prefer to denote by $\hat f$. Using (2) as reference, you can intuitively remark that $f$ is "not so different" from $g$, so $\hat f$ also should not be too different from $g$, that is it should be concentrated around zero, symmetric and have tails when $|\omega|\to\infty$. From another point of view, $f$ is concentrated, so it is somewhat a slightly enlarged $\delta$ function, and using (1) as reference you can guess that $\hat f$ will be spread over the real line and oscillate.
You can build an intuition based on remarkable correspondences like (1) and (2). Use the one we have discussed as a reference, it is a good example. $$ \Pi \;\;\underset{\mathcal F}\longleftrightarrow \;\;x\mapsto\frac{\sin x}x\tag{3}$$ ($\Pi$ is the gate function that you called $f$). Another useful correspondence is the following $$x\mapsto \exp\left(-|x|\right)\;\;\underset{\mathcal F}\longleftrightarrow \;\;x\mapsto\frac{2}{1+x^2}\tag{4},$$ it transforms a sharp peek at zero into a smooth, widespread distribution with no mean (and vice-versa). Note that in (4), no members of the pair oscillate, as in (2).
Elementary operations on Fourier transforms are also useful to understand the duality, like $$f(\lambda x) \;\;\underset{\mathcal F}\longleftrightarrow \;\;\frac{1}{\lambda}\;\hat f\!\left(\frac{x}\lambda\right) \tag{5},$$ by which I represent that dilation in a domain is associated with shrinking in the dual domain.
Another intriguing feature, that can be denoted by $$\times \underset{\mathcal F}\longleftrightarrow \otimes,\tag{6}$$ means that the Fourier transform of a product is a convolution (and vice-versa). To understand this, it is rather useful to have a good understanding of the convolution itself. Such operations are less easily pictured, but in combination with (1) they give an interesting result called the translation relation. Namely $$ \operatorname{expi} \times \underset{\mathcal F}\longleftrightarrow \;\;\text{translation}.\tag{7}$$ The translation is an operator $T$ defined by $T(f)(x)=f(x-x_0)$.
A step beyond all this is to understand the derivation operation $$ \partial\;\; \underset{\mathcal F}\longleftrightarrow \quad x\times.\tag{8}$$ Actually, (8) can be seen as a consequence of (7) using infinitesimal translation operators and the definition of the derivative as a limit.
I hope this will help you to "understand" a bit more the meaning of Fourier transform in a practical way.