So that this question has an answer:
- Observe that
$$ \|f\|_p^p = \int |f|^p. $$
Often, all we want to know is whether or not $f$ is an element of $L^p$. This happens when $\|f\|_p < \infty$, which is equivalent to $\|f\|_p^p < \infty$ (if the $L^p$-norm is finite, then the $p$-th power of that norm must also be finite). Because of this, you will often see the two values used more-or-less interchangeably.
- $|f|^p$ does not represent the $p$-fold composition of $f$ with itself. If you want to be explicit about the notation,
$$ \|f\|_p := \int_{X} |f(x)|^p \,\mathrm{d}\mu(x), $$
where $f : X \to \mathbb{C}$ (or, if you prefer to work with real valued functions, $f:X\to\mathbb{R}$) is a $\mu$-measurable function, $|f(x)|$ represents the modulus (or absolute value) of $f(x)$ so that $|f(x)|^p$ is that modulus raised to the $p$-th power, and $\mu$ is the measure on $\mathbb{X}$.
Conceptually, it might help to have a basic, basic example in mind. $L^p$ spaces generalize some notions that should already be fairly familiar to you, namely norms in finite dimensional vector spaces over $\mathbb{R}$. To get started, let's work with $\mathbb{R}^2$. In $\mathbb{R}^2$, there are a whole bunch of interesting norms that we can work with: for any $p \in [1,\infty)$, define the $p$-norm by
$$ \|\langle x,y\rangle\|_p := \sqrt[p]{|x|^p + |y|^p} = \left( |x|^p + |y|^p \right)^{1/p}. $$
There is also a reasonable $\infty$-norm, but let's not worry about that for now. Notice that if $p=2$, we recover the usual Euclidean norm, and if $p=1$, we get the "taxicab metric". It is not to hard to show that we actually get an honest-to-goodness norm for any value of $p \in [1,\infty)$. It is also worth noting that all of these norms are equivalent, in that they generate the same topology.
Generalizing this just a little bit, we can do the same thing in $\mathbb{R}^d$: if $x = \langle x_1, x_2, \dotsc, x_d \rangle \in \mathbb{R}^d$ and $p\in [1,\infty)$ we can define the $p$-norm of $x$ by
$$
\|x\|_p
:= \left( |x_1|^p + |x_2|^p + \dotsb + |x_d|^p \right)^{1/p}
= \left( \sum_{j=1}^{d} |x_j|^p \right)^{1/p}. $$
Again, it is not too difficult to show that this will always define a norm on $\mathbb{R}^d$, and that all of these norms are equivalent.
One step more general: let $x = (x_1, x_2, \dotsc) = (x_j)_{j\in\mathbb{N}}$ be a sequence with entries in $\mathbb{R}$. If we think of this sequence as being an infinite dimensional vector, then we might reasonably attempt to define $p$-norms by extending the ideas above:
$$
\|x\|_p := \left( \sum_{j=1}^{\infty} |x_j|^p \right)^{1/p}.
$$
This is entirely reasonable, and actually works! However, there are some details that change in this infinite dimensional setting: first off, it is now possible for our objects to have infinite norms. Also, for different values of $p$, the topologies on the space of sequences will be different. This isn't too hard to check, and is worth fiddling around with.
But we can phrase all of this in the language of functional analysis without too much difficulty: a sequence $(a_j)$ is really a function $a : \mathbb{N} \to \mathbb{R}$ defined by (an abuse of notation to be) $a(j) = a_j$. On this space, counting measure is natural, so what we have really done is defined
$$
\|a\|_p
= \left( \sum_{j=1}^{\infty} |a_j|^p \right)^{1/p}
= \left( \int_{\mathbb{N}} |a(j)|^p \,\mathrm{\mu}(j) \right)^{1/p},
$$
where $\mu$ is counting measure on the natural numbers. In other words, we have defined a norm on the space $(\mathbb{N}, \mathscr{P}(\mathbb{N}), \mu)$. If we restrict ourselves to functions that have finite norms, we obtain the "little ell pee" space
$$ \ell^p(\mathbb{R})
= L^p((\mathbb{N},\mu),(\mathbb{R},m))
= \left\{ a : \mathbb{N}\to\mathbb{R} \ \middle|\ \|a\|_p < \infty \right\}, $$
where $m$ is Lebesgue measure on $\mathbb{R}$.
We can now generalize to functions $f : \mathbb{R}\to \mathbb{R}$: simply replace a sequence $a : \mathbb{N} \to \mathbb{R}$ with a measurable function $f : \mathbb{R^d} \to \mathbb{R}$ and replace counting measure in that last integral with Lebesgue measure. Since $f(x)$ will be a real number for any $x\in\mathbb{R}^d$, the absolute value of $f(x)$ is well-defined, and the function $x \mapsto |f(x)|$ will be measurable (composition of measurable functions blah blah blah). Replacing the codomain $\mathbb{R}$ with $\mathbb{C}$ gives us a slightly different space to work in, but the basic theory is the same.
