I will answer this question with four typical examples, all different in nature:
Example 1:
Prove by induction that $n^3-n+3$ is divisible by $3$ $\forall \space n \in \mathbb{N^+}$.
For proof by induction; these are the $\color{red}{\mathrm{three}}$ steps to carry out:
Step 1: Basis Case: For $n=1 \implies n^3-n+3= 1^3-1+3=3$ which is divisible by $3$. So statement holds for $n=1$.
Step 2: Inductive Assumption: Assume statement is true for $n=k$ such that $n^3-n+3=\color{blue}{(k^3-k+3)}=\color{blue}{3p} \tag{1}$ Where $p,k \in \mathbb{N^+}$.
Step 3: Prove Statement holds for $n=k+1$ such that $$n^3-n+3=(k+1)^3-(k+1)+3$$
$$=k^3+3k^2+3k+1-k-1+3=3k^2+3k+\color{blue}{(k^3-k+3)}= 3(k^2+k)+\color{blue}{3p}=\color{#180}{3}(k^2+k+p)$$ using our inductive assumption $(1)$ the resulting expression clearly is divisible by $3$.
Hence $n^3-n+3$ is divisible by $3$ $\forall \space n \in \mathbb{N^+}$
QED.
(QED is an abbreviation of the Latin words "Quod Erat Demonstrandum" which loosely translated means "that which was to be demonstrated". It is usually placed at the end of a mathematical proof to indicate that the proof is complete). Alternatively, you can use $\fbox{}$
Example 2:
Use trigonometric identities and induction to prove that
$\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{n} =
\left(\begin{array}{cc}
\cos (n \theta) & -\sin (n\theta)\\
\sin (n \theta) & \cos (n \theta)
\end{array} \right)$
As before, for proof by induction; these are the $\color{red}{\mathrm{three}}$ steps to carry out:
Step 1: Basis Case: For $n=1 \implies$ LHS
$=\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{n}$
$=\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{1}$
$=\left(\begin{array}{cc}
\cos (1 \theta) & -\sin (1\theta)\\
\sin (1 \theta) & \cos (1 \theta)
\end{array} \right)$
$=\left(\begin{array}{cc}
\cos (\theta) & -\sin (\theta)\\
\sin ( \theta) & \cos ( \theta)
\end{array} \right)=$ RHS. So statement holds for $n=1$.
Step 2: Inductive Assumption: Assume statement is true for $n=k$ such that
$\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{n}$
$=\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{k}$
$=\left(\begin{array}{cc}
\cos (k \theta) & -\sin (k\theta)\\
\sin (k \theta) & \cos (k \theta)
\end{array} \right)\tag{1}$
Step 3: Prove Statement holds for $n=k+1$ such that
$\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{n}$
$=\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{k+1}$
$=\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{k}$ $\times
\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)^{1}$
$=\left(\begin{array}{cc}
\cos (k \theta) & -\sin (k\theta)\\
\sin (k \theta) & \cos (k \theta)
\end{array} \right)
\times
\left(\begin{array}{cc}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta
\end{array} \right)$
[using our inductive assumption $(1)$]
$=\left(\begin{array}{cc}
\cos\theta\cos (k \theta)-\sin\theta \sin(k\theta) & -\left(\sin \theta\cos (k \theta)+\sin (k \theta)\cos \theta\right)\\
\cos\theta\sin (k \theta)+\sin\theta \cos(k\theta) & \cos\theta\cos (k \theta)-\sin\theta \sin(k\theta)
\end{array} \right)$
$=\color{blue}{\left(\begin{array}{cc}
\cos (\theta(k+1)) & -\sin (\theta(k+1))\\
\sin (\theta(k+1)) & \cos(\theta(k+1))
\end{array} \right)}$
Where in the last step I used the fact that $$\cos(A \mp B)=\cos A \cos B \pm \sin A \sin B$$ and $$\sin(A \pm B)=\sin A \cos B \pm \cos A \sin B$$ and $\color{blue}{\mathrm{this}}$ is the same result if $n=k+1$ is substituted into the RHS of your original disposition.
Hence statement is true for all $n \in \mathbb{N}.$
QED.
Example 3:
Prove by induction that $3^{(3n+4)} + 7^{(2n+1)}$ is divisible by $11$ for all natural numbers $n$:
Step 1: Basis Case: for $n=1$: $P(1)= 3^{(3(1)+4)} + 7^{(2(1)+1)} = 2530$, which is divisible by $11$ so statement holds for $n=1$.
Step 2: Inductive Assumption: Assume statement is true for $n=k$: $P(k) =3^{(3k+4)} + 7^{(2k+1)} = 11a \implies 3^{3k+4} = \color{blue}{11a - 7^{2k+1}}$, where $a \in \mathbb{N}$
Step 3: Prove Statement holds for $n=k+1$:
$P(k+1)= 3^{3k+7} + 7^{2k+3} = 27 \cdot 3^{3k+4} + 49\cdot 7^{2k+1}\tag{1}$
Using the inductive assumption shown in $\color{blue}{\mathrm{blue}}$, $(1)$ becomes:
$27(\color{blue}{11a - 7^{2k+1}})+49\cdot 7^{2k+1}=27\cdot 11a -27\cdot 7^{2k+1}+ 49\cdot 7^{2k+1}=27\cdot 11a +2\cdot 11\cdot 7^{2k+1}=\color{red}{11}(27a+2\cdot 7^{2k+1})$
Hence $3^{3n+4} + 7^{2n+1}$ is a multiple of $\color{red}{11} \space\forall \space n\in \mathbb{N}$
QED.
Example 4:
Prove by induction that $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} \quad \forall \space n \in \mathbb{N}$$
Step 1: Basis Case: For $i=1$: $$\sum^{i=k}_{i=1} i^2=\frac{1(1+1)(2\times 1+1)}{6}= \frac{2\times 3}{6}=1$$ So statement holds for $i=1$.
Step 2: Inductive Assumption: Assume statement is true for $i=k$:
$$\sum^{i=k}_{i=1} i^2=\frac{k(k+1)(2k+1)}{6} $$
Step 3: Prove Statement holds for $i=k+1$. You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^2=\color{blue}{\frac{(k+1)(k+2)(2k+3)}{6}}$$
To do this you cannot use: $$\sum^{i=k}_{i=n} i^2=\color{red}{\frac{n(n+1)(2n+1)}{6}} $$ as this is what you are trying to prove.
So what you do instead is notice that:
$$\sum^{i=k+1}_{i=1} i^2= \underbrace{\frac{k(k+1)(2k+1)}{6}}_{\text{sum of k terms}} + \underbrace{(k+1)^2}_{\text{(k+1)th term}}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{(k+1)\left(k(2k+1)+6(k+1)\right)}{6}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{(k+1)(2k^2+\color{green}{7k}+6)}{6}=\frac{(k+1)(2k^2+\color{green}{4k+3k}+6)}{6}=\frac{(k+1)\left(2k(k+2)+3(k+2)\right)}{6}=\color{blue}{\frac{(k+1)(k+2)(2k+3)}{6}}\quad \forall \space k,n \in \mathbb{N} \quad\fbox{}$$
Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.
Observe that in the part marked $\color{green}{\mathrm{green}}$ $7k$ has simply been rewritten as $4k+3k$. From then on you simply take out common factors.
Note that this method is only valid when you have two numbers whose product is $12$ and sum is $7$.
Or, put in another way for the general quadratic $ax^2 +bx +c$, this inspection method is only valid iff you can find two numbers whose product is $ac$ and sum is $b$.
Best Answer
Think of well-formed formulae as being objects "inside the system", and the "⊢" and "," as being "outside the system" with the meaning that from the objects (inside the system) on the left we can "derive" the objects on the right. So "$\Rightarrow$" and "$\wedge$" or whatever it is that the system allows in well-formed formulae is different simply because it is inside. The rules outside the system govern how you can derive new objects in the system from previously derived or given objects in the system. So for example one typical rule is "$A$ , $B$ ⊢ $A \wedge B$ for any well-formed formulae $A$ and $B$", which means that given two objects $A$ and $B$ we can get a third object $A \wedge B$.
This is also why we need two special rules in most formal systems, one to instantiate a derived formula in the current context/scope, and another to eliminate a contradiction, both of which should not actually be written using "⊢", otherwise it is a circularly defined rule. I've seen the second written as "$( A ⊢ ( B \wedge \neg B) ) ⊢ \neg A$", but it is in my opinion a very bad practice, since a consistent interpretation of the syntax would then require "⊢" to be inside the system, which would be self-referential. Furthermore, it obscures the existence of the context/scope, which is not good. Likewise I've seen the first rule completely omitted from discussion at all! To get rid of the context/scope, one way is to use a sequent-calculus type of presentation (see http://en.wikipedia.org/wiki/Sequent_calculus), which is suitable for automated provers but not really for humans.
Now for your question of why there should be a difference at all, there are two possible reasons I know that people may give. One is that the "language" used outside the system to describe the rules of the system can be reused to describe different structures that may not be logic systems or may not behave in the same way. But beyond that, I don't see why there cannot be a single self-referential "language" that is sufficiently flexible to be used to describe everything possible. After all, we managed to use some natural language like English to successfully convey what all our mathematical symbols mean...
The second reason is to try to limit the system so that it cannot "talk about itself", and so that we can prove certain meta-theorems about the system. For example we can prove a theorem that logical equivalence of statements in boolean algebra (only logical connectives and no quantifiers) is decidable. Note that this theorem is outside the system and even requires induction, which means that we would need meta-integers and a meta-axiom for meta-induction! But what are the rules for a meta-theory for these meta-theorems? That would be a meta-meta-theory... And so on to infinity. Wait, meta$^\infty$-infinity. So I personally don't see any fundamental reason to accept this reason either.