Suppose you have some linear algebra background. The most important
thing you need to know is that the inner product has the same meaning
of what you have learnt in linear algebra class. The inner product
$$\left\langle \phi|\psi\right\rangle =\int\phi^{*}(x)\psi(x)dx$$
has the meaning related to a projection of one vector onto
another vector (for true projection, the wavefunctions needed to be normalized). It is similar to the projection of a three dimensional
vector $\mathbf{\vec{v}}=a\hat{\mathbf{x}}+b\hat{\mathbf{y}}+c\hat{\mathbf{z}}$ onto another unit vector $\mathbf{\hat{x}}$ which gives you the results $\mathbf{\vec{v}}\cdot\mathbf{\hat{x}}=a$.
First, the inner product can give you the "length square" of the wavefunction:
$$\left\langle \psi|\psi\right\rangle =\int\psi^{*}(x)\psi(x)dx =\int|\psi(x)|^2dx$$
similar to the $\mathbf{\vec{v}}\cdot\mathbf{\vec{v}}=a^{2}+b^{2}+c^{2}$
, so you can normalize your wavefunction by the condition $\left\langle \psi|\psi\right\rangle =1$.
Second, it allows you to show that two wavefunctions are orthogonal
to each other, given by the condition that the inner product evaluated
to zero $\left\langle \phi|\psi\right\rangle =0$ which is the analog
to the $\mathbf{\hat{x}}\cdot\mathbf{\hat{y}}=0$.
Third, if we write the wavefunction $\psi(x)$ as a linear combination
of orthonormal wavefunctions $\psi_{n}(x)$:
$$\psi(x)=\sum_{n}c_{n}\psi_{n}(x)$$
similar to a general vector in linear algebra, then we will have the inner
product $\left\langle \psi_{n}|\psi\right\rangle =c_{n}$. The meaning
of $c_{n}$ is the probability amplitude and it is a complex number in general.
So the probability $p_{n}$ of the wavefunction $\psi$ having the component $\psi_{n}$ is given by $p_{n}=|c_{n}|^{2}=|\left\langle \psi_{n}|\psi\right\rangle |^{2}$. The meaning here is very important when you learn how to preform measurement.
Lastly, you should add two wavefunctions amplitude together before you take the square, similar to adding the amplitude of two water waves. More precisely, if the new wavefunction is $\psi(x) = A[\psi_{a}(x)+\psi_{b}(x)]$, then the probability density at position $x$ is $A^2|\psi_{a}(x)+\psi_{b}(x)|^{2}$. Note that $A$ is the normalization constant given by the condition $\left\langle \psi|\psi\right\rangle=1$. It is where the quantum effect arise. Dont take the square and then add them together.
