Enough Voltage but No current
The internal resistance of the battery plays a larger role in the maximum current it can provide.

Suppose you have the circuit shown below:


The boxed area represents the battery and the resistor RL represents the load you are trying to power. Inside the battery, you have a voltage source and the internal resistance of the battery, which may be in the range of milliohms or less. Now imagine RL was very large, say 100kOhms. Then the total resistance of the circuit would be dominated by RL. Using Ohm's law, we can calculate the current through the circuit as
$$ I = \frac{V_o}{R_s+R_L} $$
Since Rs is much smaller than RL, we can say
$$ I = \frac{V_o}{R_L} $$
This means that the battery can supply any current that RL allows, as long as RL is much larger than Rs.

But now suppose RL is of the same order of magnitude as Rs, or even less than Rs. Then We can no longer ignore Rs in the current equation. Let's take the extreme case that RL is much smaller than Rs. Then we can ignore RL in the current equation, and we get
$$ I = \frac{V_o}{R_s} $$
This is the maximum current that the battery can supply (we essentially short the terminals of the battery together). This is because Rs is internal to the battery and we cannot change it. 

In the case of a 12V car battery, a huge current is needed to start the car's engine. Thus, the internal resistance of the battery needs to be small. A brief internet search says that the internal resistance for some car batteries is roughly 20milliOhm. This would correspond to a max current (short circuit current) of 12/0.02=600A -- the current needed to start a car engine. 

Now take an ordinary 9V battery. Again, a brief internet search says that 9V batteries have internal resistances somewhere around 1 to 2Ohm. Let's take 1Ohm. The max (short circuit) current a 9V can provide is 9/1=9A -- way less than the car battery. 



Another thing to consider is that as you draw more current from a battery (by decreasing the load resistance), more voltage will be dropped across the internal resistance, Rs. The total voltage drop across the internal resistance is again given by Ohm's law:
$$ V_s = V_0/R_s $$
The net effect of this is that the total voltage you'll see across the terminals of the battery will drop as you draw more current from the battery. At some point, the voltage will reach zero -- this is when you short the battery terminals together.