Does $f = (c + v) /$wavelength? If I'm heading at some given velocity into an electromagnetic wave with some given wavelength.
From my POV it would appear as if the wave has sped up towards me and as a consequence the wavelength of the wave would appear to decrease and hence the frequency would increase.
As the frequency of an electromagnetic wave is f = c / wavelength - to describe this scenario in an equation is all I have to do is add my velocity to the speed of light or am I missing something as I'm dealing with a wave moving at the speed of light?

 A: 
am I missing something as I'm dealing with a wave moving at the speed of light?

You are missing the second postulate of Special Relativity.
When it comes to light, it's velocity is a constant that is independent of the motion of the source or observer.
This is one of the postulates of special relativity. We cannot add velocities in the same way we do in classical mechanics when we are dealing with light.
Your equation, which would be equivalent to stating
$$c'= c+v$$ (where $v$ is your speed and $c'$ is the proposed resultant speed) is not correct. In fact $$c+v \rightarrow c$$
regardless of the value of $v$. That is, the light does not "speed up" due to your motion and you do not need to add your velocity.
It is counter-intuitive, but this is an experimentally confirmed result from the theory of relativity.
In special relativity, the relative motion of a source and an observer does cause a change in frequency/wavelength of light, but certainly not the way you described. The relativistic Doppler effect explains this by taking into account certain effects of special relativity (certainly, the constancy of the speed of light is at the heart of these effects and special relativity itself).
As you will see from the link above, the correct formulation is, for the relativistic longitudinal Doppler effect (source and observer are moving toward/away from each other)
$$\frac{f_s}{f_r} = \sqrt{  \frac{1 + \beta}{1- \beta}     }$$
is called the Doppler factor of the source relative to the receiver and $f_s$ is the frequency of the source while $f_r$ is that for the receiver, and $$\beta = \frac{v}{c}$$ You obtain an almost identical relation  for wavelength
$$\frac{\lambda_r}{\lambda_s} = \sqrt{  \frac{1 + \beta}{1- \beta}     }$$
since (as stated earlier) $c$ is constant regardless of the motion of the source or receiver. You will also note that for the transverse Doppler effect (source moving in a line transverse to observer)
$$f_r = \gamma f_s$$
and when the receiver is looking at a direct right angle to the path of the source $$f_r = \frac{f_s}{\gamma}$$ and $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}  }}$$
