Is the circuit a voltage doubler or not?
wouldn't C1 as well get charged through the resistor...[?]


You are correct that current would flow through the resistor and then through C1.  C1 would actually be discharged by this action, but by being in the circuit, its voltage (on the - side of the resistor) is added to the voltage of the power supply (D1 to the + side), resulting in double the voltage across the resistor (which is the load). C1 should not discharge all the way; if it does, then there is too much load current and the voltage ripple would be nearing 100%.  And so you would want to use a bigger capacitor.  The same action applies to C2 during the negative cycle.

I've never seen a full-wave doubler drawn this way, but of course it is correct. I slapped my head when I saw this.  And for your benefit, you should search the Web for a full-wave doubler, and look at the images, and see how it's usually drawn. It should make more sense that way.