So, I'll answer the theoretical part of your question, since we need a key to address the practical part.
Why is padding used in CBC?
Blockcipher such as AES are encrypting blocks of a fixed given size only, we call it the "blocksize". So, what if your data is smaller than the blocksize ? An easy solution is to add what we call "padding" to your plaintext in order to have its size match the block size.
So, padding is there to allow you to encrypt data smaller than the blocksize... But what if you want to encrypt something bigger than your blocksize?
That's where the mode of operations, such as CBC, come into play. Modes of operations are there to allow you to encrypt more data than the blocksize of your symmetric blockcipher.
But since those modes of operation are still using the underlying blockcipher to encrypt data, it means that they can only encrypt a multiple of the blocksize. So, if you have $4\ell$ bits of data to encrypt, for $\ell$ the block size of your cipher, then you're fine.
But what if you have $4\ell+5$ bits of data? (I am assuming the blocksize does not divide 5.) Then you need to pad the data with $\ell-5$ bits of padding in order to end up with a multiple of the blocksize of data ready for encryption with your mode of operation.
There are multiple methods to pad data, which can completely undermine the compatibility of a given encryption program with others. The most used padding method in symmetric ciphers and mode of operation is certainly PKCS#7.
There exists methods to avoid padding, in order to avoid padding completely so the ciphertext is the same size as the plaintext, the most common of those methods is certainly the one called ciphertext stealing.
There also exists modes of operation, such as CTR, which do not need any padding.
Now, contrarily to RSA encryption for instance, padding in symmetric encryption does not add security.
Checking the padding upon decryption can allow to catch errors in the transmitted ciphertext, because of the avalanche effect, (depending on the mode of operation used), but can also lead to terrific padding oracle attacks.
As explained in that question, it is better to use a cryptographic MAC to protect your ciphertext, and verify it first before performing decryption or even padding checks. I also refer you to that legendary question, regarding the way one should MAC its data when using encryption.
Actually ECB also encrypts only multiples of the blocksize, so you will also need padding or another method to have enough data to encrypt. However, you should not use ECB for the reasons pointed out there.
Regarding your practical problem
So, I've tried your values:
from Crypto.Cipher import AES
key = binascii.unhexlify("10a58869d74be5a374cf867cfb473859")
iv = binascii.unhexlify(16*"00")
plaintext = binascii.unhexlify(16*"00")
ciphertext = binascii.unhexlify("6d251e9044b051e04eaa6fb4dbf78465")
ciph = AES.new(key, AES.MODE_CBC, iv)
test1 = ciph.encrypt(plaintext)
print("The ciphertext is what we expected:", test1==ciphertext)
#We need to reset the cipher to get the correct IV
ciph = AES.new(key, AES.MODE_CBC, iv)
test2 = ciph.decrypt(ciphertext)
print("The recovered plaintext is what we expected:", test2 == plaintext)
and as you can see if you try it with Python and PyCrypto, it works but your ciphertext strangely differ from the actual value at offset 4, as pointed in the comment by fgrieu. The actual value is 69, not 90.
This error at offset 4 explains why you couldn't get the correct recovered plaintext, because of the avalanche effect, the whole AES block was being impacted by that little mistake.
8A05FC5E095AF4848A08D328D3688E3Das a ciphertext... You can test your AES-CBC using the test vectors one can find all over the internet. For example here. $\endgroup$