Michael Hardy and Johannes Kloos have answered this already, but I thought a typical example might illuminate the point further. $\def\ra{\rightarrow}$
Let's consider propositional logic and say that $S$ is some formula of propositional logic, perhaps $p\ra(p\ra q)\ra q$ or $p\ra q$. We can make a truth table of the values of $S$ for each possible valuation of its variables. If the truth table says that $S$ is true regardless of the values we assign to the variables, then we will say that $S$ is a tautology, and write $\vDash S$.
Now let's consider the following system of axioms: $ X\ra (Y\ra X) \\ (X\ra(Y\ra Z))\ra((X\ra Y)\ra(X\ra Z)) $
where $X$, $Y$, and $Z$ may be any well-formed formulas, and the modus ponens deduction rule, which says that if we can prove $X$ and we can prove $X\ra Y$, then we can prove $Y$.
If we can prove some sentence $S$ from these axioms and this deduction rule, we will say that $S$ is a theorem, and write $\vdash S$.
This would be a typical use of $\vdash$ and $\vDash$.
It should be clear that it is not at all obvious that $\vdash$ and $\vDash$ mean the same thing; they are defined quite differently. Perhaps you can imagine trying to prove that they mean the same thing.
In fact as I defined them, they do not mean the same thing! It is the case that $\vDash p\vee\lnot p$, but I gave no axioms that would allow one to deduce $\vdash p\vee\lnot p$. Every deducible theorem has a $\ra$ in it somewhere!
Or to take a less silly example, it turns out that $\vDash ((p\ra q)\ra p)\ra p$ but not $\vdash ((p\ra q)\ra p)\ra p$.
As it happens, we have $\vdash T$ implies $ \vDash T$, but not necessarily the converse. If we add $(p\ra q)\ra p)\ra p$ to the set of axiom schemas, and agree that $\vDash T$ applies only to formulas whose only operator is $\ra$, then the axiom system becomes stronger and there are fewer tautologies, and $\vdash T$ and $ \vDash T$ become equivalent.