Why can't eavesdropper not make up own private key on diffie hellmann? I try to understand diffie hellmann and I wonder why an attacker just can't make her own secret key and go along with the ride? 

She'd be ending up with just the same shared key.

Let the private keys be:
$$sk_A = 14, sk_B=12, sk_E=8$$

Let's assume the base is $g=2$ and the prime is $p=17$.

Alice would have a public key of $pk_A=13$ ($g^{sk_A} \bmod p = 2^{14} \bmod 17 = 13$) and publish it and Bob would have a public key of $pk_B=16$ ($g^{sk_B}\bmod p = 2^{12} \bmod 17 = 16$). Eavesdropper would end up with $pk_E=1$ ($g^{sk_E}\bmod p = 2^{8} \bmod 17 = 1$).

Now Alice can take the public key of Bob to arrive at the shared secret ($pk_B^{sk_A}\bmod p = 16^{14} \bmod 17 = 1$) and Bob can do so as well.

But can't the eavesdropper not always do the same procedure with her private key? In other words won't $g^{sk_E} \bmod p$ always lead to the same shared key?

(I updated the numbers so that prime is bigger than the exponents).
Eve does not have the same information. It's only with you poor choice of numbers did you get the phenomena where the number you chose for Eve happens to match. 
In general this doesn't happen, and with careful choice of parameters and large numbers the chance for this happening is negligible. 

You chose an unsafe prime $17=2^4+1$ rather than a safe prime of the form $p=2*q+1$ when q is also prime. You then chose $g=2$ which only spans a group of size 8. $2^8=1\space mod\space 17$ rather than for instance g=3 which can produce the full range. 

For Alice and Bob it is guaranteed: $(g^{P_a})^{P_b}=(g^{P_b})^{P_a} = g^{P_aP_b}$
Since Alice and Bob aren't using Eve's public key, she can't calculate the shared key with them. 
In the vast majority of cases she will end up with a different number. 

In your example the shared key came out 1, and many guesses by Eve work (but not all). Because the public key for Alice and Bob came out with low rank (2 and 4). Using a large safe prime and a generator will make this highly unlikely. 

It is considered very difficult to calculate $g^{ab}$ from $g^a$ and $g^b$ without knowing either a or b. Eve guessing a random key isn't likely to help. For random $e$ in almost all cases $g^{ae}\ne g^{ab}$