"Only dealt with vectors", you're dealing with vectors here too! Functions are vectors, and this is an inner product on a vector space!

Really, the integral is exactly the same thing as with the dot product. For two vectors in $\Bbb{R}^n$, the dot product is $(x_1,...,x_n)\cdot (y_1,...,y_n)=x_1y_1+\cdots+x_ny_n$.

For functions, you can think of the dot product being the same thing! You multiply the two components and add them up! The only difference is the measure or "weight" of each point. Since $\Bbb{R}^n$ is discrete, each component has weight $1$, where in these function spaces each component has weight "$dx$".

So loosely, on $\Bbb{R}^n$:

$$(x_1,...,x_n)\cdot (y_1,...,y_n)=x_1y_1 \times1+\cdots+x_ny_n\times 1$$

On a sequence space:

$$(x_1,...)\cdot (y_1,...)=x_1y_1 \times1+\cdots$$

On these function spaces (I write $f\cdot g$ to emphasise that I mean the dot product between the *functions* $f$ and $g$ and not the product $f(x)g(x)$ between the numbers $f(x)$ and $g(x)$):

$$f \cdot g=f(a)g(a)\times dx+f(a+dx)g(a+dx)\times dx+\cdots=\int_a^b f(x)g(x) dx$$

The inner product on function spaces is exactly the regular dot product, just in infinite dimensions and with a different "weight".

Edit, for the issue of orthogonality. The pressing issue here is that inner products define what it means to be orthogonal. So I issue a challenge to you here. Without referencing an inner product (this includes angles, as the inner product DEFINES angles), what does it mean to be orthogonal? You can't answer this question. The entire notion of angle, orthogonality, etc. are summarized in:

$$(x_1,...,x_n)\cdot (y_1,...,y_n)=x_1y_1 \times1+\cdots+x_ny_n\times 1$$

Even visualizing this is difficult. What do orthogonal vectors look like in 4 dimensions for example? I certainly don't know. Could you try picturing this in infinite dimensions? I can't.

So what does it mean to say that the inner product of two functions is $0$? The same thing it always does, that those two functions are orthogonal! Is there a nice visualization of this? Not really.

For another example, with this inner product comes the norm $\|f(x)\|_2=\sqrt{\int_a^b f^2(x) dx}$. Can you draw a unit circle? That is, all functions $f:[a,b]\to\mathbb R$ such that $\|f(x)\|_2=\sqrt{\int_a^b f^2(x) dx}=1$?

Certainly you can't. Things become less geometric in infinite dimensions.